A Family of Tetravalent Half-Transitive Graphs

arXiv:2008.07525v1 [math.CO] 16 Aug 2020 a o eactastv,tesals xml en h otgraph Holt the being called example are smallest graphs the vertex-tran Such both arc-transitive, is be which not graph may a However, edge-transitive. and rifls a:tefis st hrceiehl-rniiegraph half-transitive characterize to is majo first two the like classification, tetravalent complete of far: of classification so absence the In numerofruitful years, Though complete. 50 yet [9]. not last Holt is the in given in was published example half-tran been smallest tetravalent half-transitiv the of 2-re and a example [2] connected finding first Bouwer of any The possibility Since graph. tetravalent first or the degree. arc-transitive, even is of cycle is graph half-transitive rp hoy n srfre o[8]. to referred is one theory, graph ttrsotta ti aiyo alygah fodr3 order of graphs Cayley of family is it that out turns It [3],[13]. graphs half-transitive rpitsbitdto submitted Preprint [email protected] h uoopimgopof group automorphism the h rsof arcs the 08MSC: 2008 Keywords: aiyo ervln aftastv alygah.Aatfo th from Apart Γ( of graphs. properties Cayley structural half-transitive tetravalent of family .Introduction 1. Γ( graphs, of family new a introduce we paper, this In Abstract ∗ h td fhl-rniiegah a ntae yTte[2,who [12], Tutte by initiated was graphs half-transitive of study The nti ae,w nrdc e aiyo ervln half-transit tetravalent of family new a introduce we paper, this In graph A mi addresses: Email author Corresponding p 3 p , 4 p , G 5 G pq, , aftastv rp,gahatmrhs,cycles automorphism, graph graph, half-transitive 52,2B5 05E18 20B25, 05C25, aiyo ervln aftastv Graphs Half-transitive Tetravalent of Family A epciey ti nw hta r-rniiegahi ohvert both is graph arc-transitive an that known is It respectively. ( = eateto ahmtc,Peiec nvriy Kolkat University, Presidency Mathematics, of Department 2 pq ,E V, [email protected] t 4,5,6,7,adtescn st oeu ihifiiefmle of families infinite with up come to is second the and [4],[5],[6],[7], etc ssi ob etxtastv,eg-rniieadarc-transit and edge-transitive vertex-transitive, be to said is ) half-transitive ,a n, uhrt Biswas Sucharita G , ). Aut Agua Das) (Angsuman ( G ,at rniieyo h etcs nteegsadon and edges the on vertices, the on transitively acts ), rps o te entosrltdt algebraic to related definitions other For graphs. 1 nsmnDas Angsuman , ScaiaBiswas), (Sucharita ,a n, n .W hwta ti ninfinite an is it that show We ). n o ( for and 1, ∗ fsm atclrorders particular some of s iiegah eegvnby given were graphs sitive iieadedge-transitive and sitive prahshv been have approaches r ,India a, t edtriesome determine we at, ua sacceada and cycle a is gular rp sa4-regular a is graph e aftastv graphs half-transitive ,a n, v rps Γ( graphs, ive 9 n2 vertices. 27 on [9] (9 = ) rvdta any that proved sppr have papers us uut1,2020 19, August ex-transitive , ) eget we 4), ,a n, v if ive ). the Holt graph. In fact, our construction is a generalization of an alternative construction of Holt graph (See the last paragraph of [9]).

Definition 1.1. Let n be a positive integer such that 3 | ϕ(n), where ϕ denote the Euler Z∗ Z Totient function. Then n, the group of units of n, is a group of order a multiple of 3. Z∗ 2 Let a be a element of order 3 in n and b = a (mod n). Define Γ(n, a) to be the graph with vertex-set Zn × Z3 and the edge-set comprises of edges of the form (i, j) ∼ (ai ± 1, j − 1) and (i, j) ∼ (bi±b, j +1), where the operations in the first and second coordinates are done in modulo n and modulo 3 respectively. It is obvious that Γ(n, a) is tetravalent. One can check that Γ(9, 4) is the Holt graph. It is also to be noted that for a particular n, we can have two graphs, Γ(n, a) and Γ(n, b). However, these two graphs are isomorphic via the automorphism τ : Γ(n, a) → Γ(n, a2) defined by τ(i, j)=(ai, −j). So, without loss of generality, we assume that a < b, where a, b ∈{2,...,n − 2}. On the other hand, let n be a positive integer such that a1, b1, a2, b2 be four elements of Z∗ order 3 in n with a1b1 ≡ 1(mod n) and a2b2 ≡ 1(mod n). Then, by the above argument, Γ(n, a1) ∼= Γ(n, b1) and Γ(n, a2) ∼= Γ(n, b2). However, Γ(n, a1) may not be isomorphic to Γ(n, a2), e.g., for n = 63, we have 4·16 ≡ 1(mod 63) and 22·43 ≡ 1(mod 63), but Γ(63, 4) is not isomorphic to Γ(63, 22), as odd girth of Γ(63, 4) is 9, whereas that of Γ(63, 22) is 21. The definition of Γ(n, a) requires that 3|ϕ(n). We discuss the form of n, for which this α1 α2 αk α1−1 α2−1 αk−1 holds. Let n = p1 p2 ··· pk , wher pi are primes. Then ϕ(n)= p1 p2 ··· pk (p1− αi−1 1)(p2 − 1) ··· (pk − 1). As 3|ϕ(n), either 3|pi or 3|pi − 1 for some i, i.e., 9|n or pi ≡ 1 (mod 3) for some i. Thus n is either of the form 9t or pt where p is a prime of the form 1 (mod 3) and t is a positive integer. At this junction, it is important to mention that authors in [1] constructed a family of graphs M(a; m, n) and proved the following theorem.

Theorem 1.1 ([1], Theorem 3.3). Let n ≥ 9 be odd and a3 ≡ 1 (mod n). Then M(a;3, n) is half-transitive. While the current work was in progress, it caught our attention that our construction Γ(n, a), coincidentally, is same as M(a;3, n) (Although the deﬁnitions are given in com- pletely diﬀerent way). However for two reasons we decided to carry on with the project:

1. Firstly, the proof techniques are entirely diﬀerent: While their proof is built on semiregular automorphisms and blocks, ours is based on 6-cycles present in the graph.

2. Secondly and most importantly, we prove that Γ(n, a) is half-transitive for all n except 7 and 14, i.e., n is not necessarily odd (See Theorem 1.1). In other words, comparing with Theorem 1.1 we prove the half-transitivity of a larger family of graphs.

In Section 2, we discuss some structural properties of Γ(n, a) and in Section 3, we discuss the parameters related to the automorphism group of Γ(n, a). In Section Appendix, we provide the SageMath [11] code for computing the automorphism group of Γ(n, a).

2 2. Structural Properties of Γ(n,a) Theorem 2.1. Γ(n, a) is bipartite if and only if n is even. Proof: If n is even, then consider the sets X = {(i, j): i is even } and Y = {(i, j): Z∗ i is odd}. Note that as n is even and a, b ∈ n, therefore a, b are odd. Thus, if i is even, then ai ± 1 and bi ± b are odd, and if i is odd, then ai ± 1 and bi ± b are even. Thus, no two vertices in X (or Y ) are adjacent and hence Γ(n, a) is bipartite. On the other hand, if n is odd, as Γ(n, a) is Hamiltonian (See Corollary 3.1), it has an odd cycle of length 3n. Hence Γ(n, a) is not bipartite.

Theorem 2.2. Then chromatic number χ of Γ(n, a) is given by

2, if n is even χ = 3, if n is odd

Proof: The proof is obvious, if n is even. If n is odd, consider the sets A0 = {(i, 0) : i ∈ Zn}, A1 = {(i, 1) : i ∈ Zn} and A2 = {(i, 2) : i ∈ Zn}. The theorem is true as A0, A1 and A2 are independent sets in Γ(n, a) and Γ(n, a) is not bipartite.

Lemma 2.1. If n = p, where p ≡ 1 (mod 3) is a prime, then girth of Γ(n, a) is 3. Proof: As a3 ≡ 1(mod n) and a 6≡ 1(mod n), we have n|a3 − 1=(a − 1)(a2 + a + 1). As n is prime, we have a2 + a +1 ≡ 0(mod n). The lemma follows from the observation that (0, 0) ∼ (1, 2) ∼ (a +1, 1) ∼ (a2 + a +1, 0) = (0, 0) is a 3-cycle.

Lemma 2.2. Γ(n, a) always have a cycle of length 6 and hence girth of Γ(n, a) is less than or equal to 6. Proof: Consider the cycle (0, 0) ∼ (1, 2) ∼ (a − 1, 1) ∼ (a2 − a +1, 0) ∼ (−a2 + a, 2) ∼ (b, 1) ∼ (0, 0). Clearly it is a 6-cycle, provided the vertices are distinct. If two vertices are not distinct, then we have either a − 1 ≡ b or −a2 + a ≡ 1 or a2 − a +1 ≡ 0, i.e., in all cases, we should have a2 − a +1 ≡ 0 (mod n), i.e., a3 +1 ≡ 0 (mod n). This implies 2 ≡ 0 (mod n), a contradiction.

Lemma 2.3. Γ(n, a) does not have any 4-cycle. Proof: Let, if possible, Γ(n, a) has a 4-cycle. As Γ(n, a) is edge-transitive (by Theorem 3.3), without loss of generality, we can assume that it has (0, 0) and (1, 2) as two of its adjacent vertices. Now, the other three neighbours of (0, 0) are X = {(b, 1), (−b, 1), (−1, 2)} and that of (1, 2) are Y = {(2b, 0), (a+1, 1), (a−1, 1)}. To reach a contradiction, it suﬃces to show that X and Y does not have any edge between them. We start with (2b, 0) ∈ Y . If (2b, 0) ∼ (−1, 2), then by the adjacency criterion, we have 2ab ± 1 ≡ −1 (mod n), i.e., −1 ≡ 3, 1 (mod n), i.e., n divides 2 or 4, a contradiction. If (2b, 0) ∼ (b, 1), then by the adjacency criterion, we have ab ± 1 ≡ 2b (mod n), i.e., 2b ≡ 0, 2(mod n). Now, 2b ≡ 0(mod n) implies that b is zero divisor in Zn, a contradiction. So, let 2b ≡ 2 (mod n). If n is odd, this implies b ≡ 1(mod n), a contradiction. If n =2m is even, then this implies b ≡ 1( mod m), i.e., b = m +1 (as b =6 1). However, this in turn implies b3 = m(m2 +m+3)+1. Note that m2 +m+3 is odd irrespective of m is odd/even.

3 Thus b3 =1(6 mod 2m), a contradiction. Thus, (2b, 0) 6∼ (b, 1). Similarly, it can be shown that (2b, 0) 6∼ (−b, 1). Thus (2b, 0) is not adjacent to any vertices in X. Now, let us consider (a +1, 1) ∈ Y . The only possible neighbour of (a +1, 1) in X is (−1, 2). If(a+1, 1) ∼ (−1, 2), then we have (a+1)b±b ≡ −1(mod n). As −1 6≡ 1(mod n), we have 1 + 2b ≡ −1 (mod n), i.e., 2b ≡ −2 (mod n). But, as in previous case, this leads to a contradiction, thereby establishing that (a +1, 1) 6∼ (−1, 2). Similarly, it can be shown that (a − 1, 1) ∈ Y has no neighbour in X. Hence the lemma follows.

Corollary 2.4. If n is even, then girth of Γ(n, a) is 6. Proof: This follows from Theorem 2.1, Lemma 2.2 and Lemma 2.3.

Lemma 2.5. If 9|n, then Γ(n, a) is triangle-free. Proof: If possible, let Γ(n, a) conatains a triangle. As Γ(n, a) is edge-transitive, without loss of generality, we can assume that two of the vertices of the triangle are (0, 0) and (1, 2). Then the third vertex is of the form (x, 1). As it is adjacent to (0, 0), x ≡ ±b (mod n) and as it is adjacent to (1, 2), x ≡ a ± 1 (mod n). Combining these two, we get x ≡ a ± 1 ≡ ±b ≡ a2 (mod n). Thus, we have four cases: Case 1: a − 1 ≡ a2 (mod n). In this case, multiplying both sides by a, we have a2 − a ≡ 1 (mod n), i.e., n|2, a contradiction. Case 2: a +1 ≡ a2 (mod n). In this case, multiplying both sides by a, we have a2 + a ≡ 1 (mod n), i.e., 2a ≡ 0 (mod n), a contradiction, as a is not a zero-divisor. Case 3: a − 1 ≡ −a2 (mod n). In this case, multiplying both sides by a, we have a2 − a ≡ −1 (mod n), i.e., 2a ≡ 2(mod n). Again, as in the proof of Lemma 2.3, we reach a contradiction. Case 4: a+1 ≡ −a2 (mod n), i.e., a2 +a+1 ≡ 0(mod n). As9|n, we have 9|(a2 +a+1). Moreover, as a is a unit in Zn, we have gcd(9, a) = 1. Thus a is of the form 9k + i, where i =1, 2, 4, 5, 7, 8. Now, it can be checked for all such choices of i, a2 +a+1 is not a multiple of 9, a contradiction. Combining all the cases, the lemma holds. 5 if n =9 Lemma 2.6. Let 9|n. Then girth of Γ(n, a)= 6 if n =96 Proof: If n is an even multiple of 9, then the lemma follows from Corollary 2.4. So, we assume that n =9k where k is odd. Clearly (0, 0) ∼ (1, 2) ∼ (5, 1) ∼ (6, 2) ∼ (7, 1) ∼ (0, 0) is a 5-cycle in Γ(9, 4), the Holt graph. Thus from Lemma 2.3 and Lemma 2.5, it follows that girth of Γ(9, 4) is 5. If possible, let n = 9k where k ≥ 3 is an odd number and girth of Γ(n, a) is 5. Let C be a 5-cycle in Γ(n, a). As Γ(n, a) is edge-transitive (by Theorem 3.3), without loss of generality, we can assume that C has (0, 0) and (1, 2) as two of its adjacent vertices. Let X = {(−1, 2), (b, 1), (−b, 1)} be the other three neighbours of (0, 0) and Y = {(2a2, 0), (a+ 1, 1), (a − 1, 1)} be the other three neighbours of (1, 2). Since C is a 5-cycle, some vertex in X and some vertex in Y must have a common neighbour.

4 We will show that this is not possible. As each X and Y has 3 vertices, we need to check that all possible 9 pairs do not have any common neighbour. However, as the proof technique is similar for all the pairs, we show it only for one pair, namely (b, 1) = (a2, 1) and (a +1, 1). The common neighbour, if any, of (a2, 1) and (a +1, 1), is of the form (x, 0) or (x, 2). Case 1: Let (x, 0) be the common neighbour of (a2, 1) and (a +1, 1). Then, by the adjacency criterion, we have x ≡ a3 ±1 ≡ a2 +a±1(mod n), i.e., a2 +a±1 ≡ 0or2(mod n). If a2 + a ± 1 ≡ 0(mod‘n), then we land in Case 3 or Case 4 of the proof of Lemma 2.5 and hence a contradiction. If a2 + a +1 ≡ 2 (mod‘n), then also we land in Case 3 of Lemma 2.5 and hence a contradiction. If a2 + a − 1 ≡ 2 (mod‘n), i.e., a2 + a − 3 ≡ 0 (mod‘n), then multiplying both sides by a, we get a2 − 3a +1 ≡ 0 (mod n). Hence, subtracting on eequation from the other, we have 2(1 − a) ≡ 0 (mod n). Now, as n is odd, this implies Z∗ a ≡ 1 (mod n), a contradiction to the fact that ◦(a) = 3 in n. Case 2: Let (x, 1) be the common neighbour of (a2, 1) and (a +1, 1). Then, by the adjacency criterion, we have x ≡ 1 + a2 ± a2 ≡ a ± a2 (mod n), i.e., a ± a2 ≡ 1 or 1+2a2 (mod n). If a ± a2 ≡ 1, then we reach either Case 1 or Case 3 of the proof of Lemma 2.5 and hence a contradiction. Also, if a + a2 ≡ 1+2a2 (mod n), we reach Case 1 of the proof of Lemma 2.5 and hence a contradiction. If a − a2 ≡ 1+2a2 (mod n), i.e., 3a2 − a +1 ≡ 0(mod n), then multiplying both sides by a, we get a2 − a − 3 ≡ 0 (mod n). Subtracting one equation from the other yields 2(2a + 1) ≡ 0 (mod n). As n is odd, this implies 2a ≡ −1 (mod n). Cubing both the sides, we get 8 ≡ 8a3 ≡ −1 (mod n), which is possible only if n = 9, a contradiction. Finally, combining both the cases, the lemma holds.

3. Automorphisms of Γ(n,a) Let G = Aut(Γ(n, a)). We start by noting the following automorphisms of Γ(n, a).

α :(i, j) 7→ (i + a−j, j); β :(i, j) 7→ (i, j + 1); γ :(i, j) 7→ (−i, j);

It can be shown that α,β,γ ∈ G and ◦(α)= n, ◦(β) = 3 and ◦(γ) = 2. Moreover, we have 2 the following relations: αβ = βαa ,αγ = γα−1 and βγ = γβ. Theorem 3.1. Γ(n, a) is a Cayley graph. Proof: Let H = hα, βi. Clearly it forms a subgroup of G. Also as ◦(α)= n, ◦(β) = 3 and 2 αβ = βαa , we have

H = {αiβj :0 ≤ i ≤ n − 1, 0 ≤ j ≤ 2} and |H| =3n = |Γ(n, a)|.

We will show that H acts regularly on Γ(n, a). As |H| = |Γ(n, a)|, it is enough to show −j that H acts transitively on Γ(n, a). As i 7→ i + a is a permutation of Zn order n and j 7→ j + 1 is a permutation of Z3 order 3, the action of H on Γ(n, a) is transitive. 2 Note that H is a semidirect product of hαi and hβi, as β−1αβ = αa and a2 and n are coprime, and Γ(n, a) = Cay(H,S) where S = {β2α, β2α−1,βαb,βα−b}. We now recall a result on hamiltonicity of Cayley graphs.

5 Theorem 3.2 ([10], Theorem 3.3). Every connected Cayley graph of a semidirect product of a cyclic group of prime order by an abelain group of odd order is Hamiltonian.

Corollary 3.1. Γ(n, a) is Hamiltonian, if n is odd. Proof: By Theorem 3.1, we have Γ(n, a) is a Cayley graph on H and H is a semidirect product of cyclic group of order 3, namely hβi and another cyclic group of odd order n, namely hαi. Thus, by Theorem 3.2, Γ(n, a) is Hamiltonian.

Theorem 3.3. Γ(n, a) is edge-transitive. Proof: As Γ(n, a) is Cayley, it is vertex-transitive. Hence, it is enough to show that any two edges incident to (0, 0) can be permuted by an automorphism. As Γ(n, a) is tetravalent, the four vertices adjacent to (0, 0) are namely: (1, 2), (−1, 2), (b, 1)and (−b, 1). Let us name the following edges:

e1 : (0, 0) ∼ (1, 2) e2 : (0, 0) ∼ (−1, 2) e3 : (0, 0) ∼ (b, 1) e4 : (0, 0) ∼ (−b, 1)

← ← It is to be noted that γ(e1)= e2, αβγ(e1)= e3 and γαβγ(e1)= e4. The reverse arrow on top denote that the orientation of the edge changed. Hence, the theorem. For n =7, 14, SageMath [11] computation shows that Γ(n, a) is arc-transitive. Next we prove that Γ(n, a) is not arc-transitive if n =76 , 14. For that, we show that there does not exist any graph automorphism ϕ which maps the arc e3 to e1, i.e., ϕ((0, 0)) = (0, 0) and ϕ((b, 1)) = (1, 2). If possible, let such an automorphism ϕ exists with ϕ((0, 0)) = (0, 0) and ϕ((b, 1)) = (1, 2). As (1, 2) is adjacent to (0, 0), its image under ϕ should be one among (b, 1), (−b, 1) and (−1, 2). However, the next theorem shows that this can not hold.

Theorem 3.4. If ϕ is an automorphism of Γ(n, a) such that n =76 , 14 and ϕ((0, 0) = (0, 0) and ϕ((b, 1)) = (1, 2), then ϕ((1, 2)) ∈{/ (b, 1), (−b, 1), (−1, 2)}. Thus, from Theorem 3.1, Theorem 3.3 and Theorem 3.4, we get the desired result as follows:

Theorem 3.5. Γ(n, a) is half-transitive if n =76 , 14.

4. Proof of Theorem 3.4 To prove Theorem 3.4, we prove a lemma and three theorems. Throughout this section, ϕ denote an automorphism of Γ(n, a) and G denote the full automorphsim group of Γ(n, a).

Lemma 4.1. The following relations can not hold. 1. 2a − 4b = 0, 2. 2a +4b = 0 except n = 9, 3. 4a − 2b = 0 except n = 7, 14, 4. 4a +2b =0 except n = 18, 5. 2a − 2b =0, 6. 2a +2b =0, 7. 4a +4=0, 8. 2a +6=0, 9. 2(a + b − 1)=0, 10. 2(b − a +1)=0, 11. 2(a − b +1)=0, 12. 2(a + b +2)=0, 13. 2(a − b − 2)=0. Proof: 1. 2a − 4b, i.e., 8 = 64, i.e., 56 = 0, i.e., n | 56, hence n = 7, 14, 28, 56 and

6 the possible values of b are 4, 11, 25, 25 respectively. In all these cases 2b =6 4, which is a contradiction. 2. 2a +4b = 0, i.e., 8 = −64, i.e., 72 = 0, i.e., n | 72, hence n = 9, 18, 36, 72 and the possible values of b are 7, 13, 25, 49. But the relation holds only if n = 9 and b = 7. 3. 4a − 2b = 0, 8 = 64, i.e., 56 = 0, i.e., n | 56, hence n = 7, 14, 28, 56 and the possible values of (a, b) are (2, 4), (9, 11), (9, 25), (9, 25) respectively. But the relation holds only if n =7, 14. 4. 4a +2b = 0, i.e., 64 = −8, i.e., 72 = 0, i.e., n | 72, hence n = 9, 18, 36, 72 and the possible values of (a, b) are (4, 7), (7, 13), (13, 25), (25, 49) respectively. But the relation holds only if n = 18. 5. 2a − 2b = 0, i.e., 2a ≡ 2(mod n). If n is odd then a = 1, which is impossible. Let n be even and n = 2m. Then we have m | a − 1, i.e., a = mt + 1, for some t ∈ Z. As a =6 1 so a = m + 1, i.e., a3 − 1= m(m2 +3m + 3). Note that irrespective of m is odd or even, (m2 +3m +3) is odd, say (2s + 1), for some s ∈ Z. So we have a3 − 1= m(2s + 1), i.e., a3 − 1 ≡ m(mod n), which is a contradiction. 6. The proof is same as 5. 7. 4a + 4 = 0, i.e., 4a ≡ −4(mod n). If n is odd then a = −1, which is impossible. Let n is even and n =2m, then we have m | 2(a + 1), i.e., 2a = mt − 2, for some t ∈ Z. As2a =6 −2 so 2a = m − 2, i.e., 8(a3 − 1) = m(m2 − 6m +12) − 16. If m is even then (m2 − 6m + 12) is even, say 2u, for some u ∈ Z. So we have 8(a3−1)=2mu−16, i.e., 8(a3−1) ≡ −16(mod n), which is a contradiction. If m is odd then (m2 − 6m +12) is odd, say 2v + 1, for some v ∈ Z. So we have 8(a3 − 1) = m(2v + 1) − 16, i.e., 8(a3 − 1) ≡ m − 16(mod n), which is a contradiction as m =6 16. 8. 2a + 6 = 0, i.e., 8 = −216, i.e., 224 = 0, i.e., n | 224, i.e., n = 7, 14, 28, 56, 112, 224. However, in all these cases, the possible values of a does not allow 2a + 6 = 0. 9. 2(a + b − 1) = 0, i.e., 2(1 + a − b) = 0, i.e., 4a = 0, contradicting that a is a unit. 10. The proof is same as 9. 11. 2(a − b + 1) = 0, i.e., 2(1 − a + b) = 0, i.e., 2(a − b +1)+2(1 − a + b) = 0, i.e., 4 = 0, which is a contradiction. 12. 2(a + b + 2) = 0, i.e., 2(1 + a +2b) = 0, i.e., 4(a + b + 2) − 2(1 + a +2b) = 0, i.e., 2(a + 3) = 0, i.e., 8 = −216, i.e., 224 = 0, i.e., n | 224, i.e., n =7, 14, 28, 56, 112, 224. In all these cases 2a +6 =6 0, which is a contradiction. 13. 2(a − b − 2) = 0, i.e., 2(1 − a − 2b) = 0, i.e., 2(a − b − 2) + 2(1 − a − 2b) = 0, i.e., 6b + 2 = 0, i.e., 2a +6 =0. Rest of the proof is same as 8.

Theorem 4.1. If ϕ ∈ G and ϕ((0, 0)) = (0, 0), ϕ((b, 1)) = (1, 2), ϕ((1, 2)) = (−1, 2) then n =7 or 14. Proof: Consider the cycle C : (0, 0) ∼ (b, 1) ∼ (a + b, 2) ∼ (1 + a + b, 0) ∼ (a + 1, 1) ∼ (1, 2) ∼ (0, 0). Then ϕ(C) : (0, 0) ∼ (1, 2) ∼ ϕ((a + b, 2)) ∼ ϕ((1 + a + b, 0)) ∼ ϕ((a +1, 1)) ∼ (−1, 2) ∼ (0, 0). As ϕ((a + b, 2)) ∼ (1, 2) and ϕ((0, 0)) = (0, 0) so ϕ((a + b, 2)) ∈ {(2b, 0), (a ± 1, 1)}. Again, since ϕ((a +1, 1)) ∼ (−1, 2) and ϕ((0, 0)) = (0, 0) imply ϕ((a +1, 1)) ∈ {(−2b, 0), (−a ± 1, 1)}. Also ϕ((1 + a + b, 0)) ∼ ϕ((a + b, 2)) and

7 ϕ((b, 1)) = (1, 2) imply

ϕ((1+ a + b, 0)) ∈{(2a ± b, 1), (3, 2), (1+2b, 2), (b + a ± 1, 0), (1 − 2b, 2), (b − a ± 1, 0)}. (1)

If ϕ((a +1, 1)) = (−2b, 0) then ϕ((1+ a + b, 0)) ∼ ϕ((a +1, 1)) and ϕ((1, 2)) = (−1, 2)) imply ϕ((1 + a + b, 0)) ∈{(−3, 2), (−2a ± b, 1)} (2) From the Equations 1 and 2 we have,

• either −3 = 3, i.e., 6 = 0, i.e., n = 6 which is impossible.

• or −3=1+2b, i.e., 2a +4b = 0, which is possible only when n = 9, (by Lemma 4.1). However, direct Sagemath computation for n = 9 shows that such ϕ does not exist.

• or −3=1 − 2b, i.e., 2a − 4b = 0, which is impossible by Lemma 4.1.

• or −2a ± b =2a ± b, i.e., 4a =0or4a − 2b =0or4a +2b = 0. Though the ﬁrst one is impossible, the other two can hold only if n =7, 14, 18 (by Lemma 4.1). However, direct Sagemath computation for n =7, 14 and 18 shows that such ϕ does not exist.

Hence ϕ((a +1, 1)) =(6 −2b, 0). If ϕ((a+1, 1)) = (−a+1, 1), then ϕ((1+a+b, 0)) ∼ ϕ((a+1, 1)) and ϕ((1, 2)) = (−1, 2) imply ϕ((1 + a + b, 0)) ∈{(−1+2b, 2), (−b + a ± 1, 0)} (3) From the Equations 1 and 3 we have

• either −b + a ± 1= b − a ± 1, i.e., 2(a − b)=0or2(b − a +1) =0or2(b − a − 1) = 0, all of which are impossible by Lemma 4.1.

• or −b + a ± 1= b + a ± 1, i.e., 2b = 0 or2a − 2b =0or2a +2b = 0, all of which are impossible by Lemma 4.1.

• or −1+2b = 3, i.e., 2a − 4b = 0, which is impossible by Lemma 4.1.

• or −1+2b =1+2b, i.e., 2 = 0, which is a contradiction.

• or −1+2b =1 − 2b, i.e., 4a − 2b = 0 which can hold only if n = 7 or 14. (by Lemma 4.1.) However, direct Sagemath computation for n = 7, 14 shows that such ϕ does not exist.

Hence ϕ((a +1, 1)) =(6 −a +1, 1). If ϕ((a+1, 1)) = (−a−1, 1) then ϕ((1+a+b, 0)) ∼ ϕ((a+1, 1)) and ϕ((1, 2)) = (−1, 2)) imply ϕ((1 + a + b, 0)) ∈{(−1 − 2b, 2), (−b − a ± 1, 0)} (4) From the Equations 1 and 4 we have

8 • either −1 − 2b = 3, i.e., 2a +4b = 0, which can hold only if n = 9. (by Lemma 4.1). However, direct Sagemath computation rules out this possibility. • or −1 − 2b =1+2b, i.e., 4a +2b = 0, which can hold only if n = 18. (by Lemma 4.1). However, direct Sagemath computation rules out this possibility. • or −1 − 2b =1 − 2b, i.e., 2 = 0, which is a contradiction. • or −b − a ± 1= b − a ± 1, i.e., 2b =0, or2a +2b =0, or 2a − 2b = 0 all of which are impossible by Lemma 4.1. • or −b − a ± 1= b + a ± 1, i.e., 2(b + a − 1)=0 or 2(b + a) = 0 (which are impossible by Lemma 4.1), but 2(1 + a + b) = 0 may hold. Therefore we have ϕ((1 + a + b, 0)) = (1+ a + b, 0), ϕ((a + 1, 1)) = (−a − 1, 1), ϕ((a + b, 2)) = (a +1, 1) with 2(a + b + 1) = 0. Consider the cycle C′ : (1+ a + b, 0) ∼ (a +1, 1) ∼ (1+2b, 2) ∼ (2a, 0) ∼ (b +2, 1) ∼ (a + b, 2) ∼ (1 + a + b, 0). Then ϕ(C′) : (1+ a + b, 0) ∼ (−a − 1, 1) ∼ ϕ((1 + 2b, 2)) ∼ ϕ((2a, 0)) ∼ ϕ((b +2, 1)) ∼ (a +1, 1) ∼ (1+ a + b, 0). Now ϕ((b +2, 1)) ∼ (a +1, 1), ϕ((1+ a+b, 0)) = (1+a+b, 0) and ϕ((b, 1)) = (1, 2) imply ϕ((b+2, 1)) ∈{(1+2b, 2), (b+a−1, 0)}. Again ϕ((1+2b, 2)) ∼ (−a − 1, 1), ϕ((a + b +1, 0)) = (a + b +1, 0)=(−a − b − 1, 0) and ϕ((1, 2)) = (−1, 2) imply ϕ((1+2b, 2)) ∈{(−1 − 2b, 2), (−b − a +1, 0)}. Also ϕ((2a, 0)) ∼ ϕ((b +2, 1)) and ϕ((a + b, 2)) = (a +1, 1) imply ϕ((2a, 0)) ∈{(b +2a ± b, 0), (a +3, 1), (a +1 − 2b, 1), (1 + b − a ± 1, 2)}. (5) Let ϕ((1+2b, 2)) = (−1−2b, 2). ϕ((2a, 0)) ∼ ϕ((1+2b, 2)) and ϕ((a+1, 1)) = (−a−1, 1) imply ϕ((2a, 0)) ∈{(−b − 2a ± b, 0), (−a − 3, 1). (6) From the Equations 5 and 6 we have • either −b − 2a ± b = b +2a ± b, i.e., 4a +4=0or4a = 0 (which are impossible by Lemma 4.1) or 4a+2b = 0, which can hold only if n = 18. However, direct Sagemath computation rules out this possibility. • or −a − 3= a + 3, i.e., 2a + 6 = 0, which is impossible by Lemma 4.1. • or −a − 3= a +1 − 2b, i.e., 2(a − b +2) = 0. Also, we had 2(a + b + 1) = 0 previously. This yields 2a = 4, i.e., n = 7 or 14. Hence ϕ((1 + 2b, 2)) = (−1 − 2b, 2) is possible only if n = 7 or 14. Moreover, direct Sagemath computation for n = 7 and 14 conﬁrms the possibility. Let ϕ((1 + 2b, 2)) = (−b − a +1, 0). ϕ((2a, 0)) ∼ ϕ((1 + 2b, 2)) and ϕ((a +1, 1)) = (−a − 1, 1) imply ϕ((2a, 0)) ∈{(−a − 1+2b, 1), (−1 − b + a ± 1, 2). (7) From the Equations 5 and 7 we have,

9 • either −1 − b + a ± 1=1+ b − a ± 1, i.e., 2(b − a +1) = 0 or 2(a − b) = 0or 2(a − b − 2) = 0, all of which are impossible by Lemma 4.1.

• or −a − 1+2b = a +1 − 2b, i.e., 2(a +1 − 2b) = 0, i.e., 2(a + b − 2) = 0. Hence combining 2(a + b +1)=0 and 2(a + b − 2) = 0, we have 6 = 0, which is impossible.

• or −a − 1+2b = a + 3, i.e., 2(a − b +2) = 0. Therefore from 2(a + b +1) = 0 and 2(a − b + 2) = 0 we have 2a = 4, i.e., n = 7 or 14. Hence ϕ((1 + 2b, 2)) = (−b − a +1, 0) may be possible if n = 7 or 14. Moreover, direct Sagemath computation for n = 7 and 14 conﬁrms the possibility. Therefore for ϕ ∈ G we can have ϕ((0, 0)) = (0, 0), ϕ((b, 1)) = (1, 2), ϕ((1, 2)) = (−1, 2) only if n = 7 or 14. Theorem 4.2. If ϕ ∈ G and ϕ((0, 0)) = (0, 0), ϕ((b, 1)) = (1, 2) then ϕ((1, 2)) =(6 −b, 1). Proof: If possible let ϕ ∈ G and ϕ((0, 0)) = (0, 0), ϕ((b, 1)) = (1, 2) and ϕ((1, 2)) = (−b, 1). Consider the cycle C : (0, 0) ∼ (b, 1) ∼ (a + b, 2) ∼ (1 + a + b, 0) ∼ (a +1, 1) ∼ (1, 2) ∼ (0, 0). Then ϕ(C) : (0, 0) ∼ (1, 2) ∼ ϕ((a + b, 2)) ∼ ϕ((1 + a + b, 0)) ∼ ϕ((a + 1, 1)) ∼ (−b, 1) ∼ (0, 0). As ϕ((a + b, 2)) ∼ (1, 2) and ϕ((0, 0)) = (0, 0) so ϕ((a + b, 2)) ∈ {(2b, 0), (a±1, 1)}. Again, ϕ((a+1, 1)) ∼ (−b, 1) and ϕ((0, 0)) = (0, 0) imply ϕ((a+1, 1)) ∈ {(−2, 0), (−a ± b, 2)}. Now as ϕ((1 + a + b, 0)) ∼ ϕ((a + b, 2)) and ϕ((b, 1)) = (1, 2) then we have,

ϕ((1+ a + b, 0)) ∈{(2a ± b, 1), (3, 2), (1+2b, 2), (b + a ± 1, 0), (1 − 2b, 2), (b − a ± 1, 0)}. (8)

Depending upon the value of ϕ((a +1, 1)), one of the three cases, namely (Case A: ϕ((a+1, 1)) = (−2, 0)), (Case B: ϕ((a+1, 1)) = (−a−b, 2)) and (Case C: ϕ((a+1, 1)) = (−a + b, 2)) must hold. However, before resolving this three cases, we prove a claim which will be crucial in the following proof. Claim: ϕ((−1 − a − b, 0)) ∈{(−2a ± b, 1), (−3, 2), (−1+2b, 2), (−b + a ± 1, 0), (−1 − 2b, 2), (−b − a ± 1, 0), (2a ± 1, 2), (3b, 1), (b +2, 1), (1 + a ± b, 0), (b − 2, 1), (1 − a ± b, 0)}. Proof of Claim: As (−b, 1) ∼ (0, 0) and(−1, 2) ∼ (0, 0), we have ϕ((−b, 1)),ϕ((−1, 2)) ∈ {(b, 1), (−1, 2)}. Case 1: Let ϕ((−b, 1)) = (−1, 2) and ϕ((−1, 2)) = (b, 1). Consider the cycle C′ : (0, 0) ∼ (−b, 1) ∼ (−a−b, 2) ∼ (−1−a−b, 0) ∼ (−a−1, 1) ∼ (−1, 2) ∼ (0, 0), then ϕ(C′): (0, 0) ∼ (−1, 2) ∼ ϕ((−a − b, 2)) ∼ ϕ((−1 − a − b, 0)) ∼ ϕ((−a − 1, 1)) ∼ (b, 1) ∼ (0, 0). As ϕ((−a − b, 2)) ∼ (−1, 2) and ϕ((0, 0)) = (0, 0) so ϕ((−a − b, 2)) ∈{(−2b, 0), (−a ± 1, 1)}. ϕ((−a − 1, 1)) ∼ (b, 1) and ϕ((0, 0)) = (0, 0) imply ϕ((−a − 1, 1)) ∈ {(2, 0), (a ± b, 2)}. Now ϕ((−1 − a − b, 0)) ∼ ϕ((−a − b, 2)) and ϕ((−b, 1)) = (−1, 2) imply

ϕ((−1−a−b, 0)) ∈{(−2a±b, 1), (−3, 2), (−1+2b, 2), (−b+a±1, 0), (−1−2b, 2), (−b−a±1, 0)}. (9) Case 2: Let ϕ((−b, 1)) = (b, 1) and ϕ((−1, 2)) = (−1, 2). Consider the cycle C′ : (0, 0) ∼ (−b, 1) ∼ (−a − b, 2) ∼ (−1 − a − b, 0) ∼ (−a − 1, 1) ∼ (−1, 2) ∼ (0, 0), then

10 ϕ(C′):(0, 0) ∼ (b, 1) ∼ ϕ((−a−b, 2)) ∼ ϕ((−1−a−b, 0)) ∼ ϕ((−a−1, 1)) ∼ (b, 1) ∼ (0, 0). As ϕ((−a − b, 2)) ∼ (b, 1) and ϕ((0, 0)) = (0, 0) so ϕ((−a − b, 2)) ∈ {(−2, 0), (a ± b, 2)}. ϕ((−a−1, 1)) ∼ (−1, 2) and ϕ((0, 0)) = (0, 0) imply ϕ((−a−1, 1)) ∈{(−2b, 0), (−a±1, 1)}. Now ϕ((−1 − a − b, 0)) ∼ ϕ((−a − b, 2)) and ϕ((−b, 1)) = (b, 1) imply

ϕ((−1−a−b, 0)) ∈{(2a±1, 2), (3b, 1), (b+2, 1), (1+a±b, 0), (b−2, 1), (1−a±b, 0)}. (10)

Combining Case 1 and 2, the claim follows. Now, we turn towards the three cases mentioned earlier. Case A: If ϕ((a +1, 1)) = (−2, 0) then ϕ((1+ a + b, 0)) ∼ ϕ((a +1, 1)) and ϕ((1, 2)) = (−b, 1) imply ϕ((1 + a + b, 0)) ∈{(−3b, 1), (−2a ± 1, 2)} (11) From the Equations 8 and 11 we have,

• either −3b =2a ± b, i.e., −2b =2a or 2b = −4. By Lemma 4.1, this can hold only if n = 9. However, direct SageMath computation for n = 9 show that such ϕ does not exist.

• or −2a ± 1 = 3, i.e., −2a =4or −2a = 2, i.e., 4a +2b =0or2a +2b = 0. By Lemma 4.1, 2a +2b = 0 can not hold and 4a +2b = 0 can hold only if n = 18. However, direct SageMath computation for n = 18 shows that such ϕ does not exist.

• or −2a ± 1=1 − 2b, i.e., 2a =2b, 2(b − a − 1) = 0, both of which are impossible by the Lemma 4.1.

• or −2a ± 1=1+2b, i.e., 2a +2b = 0 (which is impossible by the Lemma 4.1) but

2a +2b +2=0 may hold. (12)

When ϕ((a +1, 1)) = (−2, 0), ϕ((1 + a + b, 0)) = (1+2b, 2), ϕ((a + b, 2)) = (a +1, 1) then we have the Equation 12. As 2a +2b + 2 = 0, i.e., a + b +1 = −a − b − 1, then ϕ((−1−a−b, 0)) = (1+2b, 2). But from the Equations 9 and 10 we have ϕ((−1−a−b, 0)) =6 (1+2b, 2), which is a contradiction. Hence ϕ((a +1, 1)) =6 (−2, 0) and Case A can not hold. Case B: If ϕ((a + 1, 1)) = (−a − b, 2) then ϕ((1 + a + b, 0)) ∼ ϕ((a +1, 1)) and ϕ((1, 2)) = (−b, 1)) imply

ϕ((1 + a + b, 0)) ∈{(−b − 2, 1), (−1 − a ± b, 0)} (13)

From the Equations 8 and 13 we have, Case B(1): either −b − 2=2a ± b, i.e., 2a +2b = 0, which is impossible by the Lemma 4.1, or 2a +2b +2=0 may hold. (14)

11 Case B(2): or we have −1 − a ± b = b + a ± 1, i.e., 2a +2b =0or2a = 0 (which are impossible by Lemma 4.1) but −1 − a − b = b + a + 1, i.e.,

2a +2b +2=0 may hold. (15)

Case B(3): or we have −1 − a ± b = b − a ± 1. This gives rise to four equations, out of which three are impossible by Lemma 4.1, namely 2 = 0, 2b = 0 and 2a +2b = 0. The only possibility which remains is −1 − a + b = b − a − 1 and it is an identity. So assuming this identity, we have ϕ((a+1, 1)) = (−a−b, 2), ϕ((a+b+1, 0)) = (b−a− 1, 0) and ϕ((a + b, 2)) = (a − 1, 1). Similarly we can show that ϕ((a − 1, 1)) = (−a + b, 2), ϕ((b−a+1, 0)) = (b+a−1, 0) and ϕ((a−b, 2)) = (a+1, 1). Now ϕ((a+b, 2)) = (a−1, 1), ϕ(((a − b, 2)) = (a +1, 1) and ϕ((2, 0)) ∼ ϕ((b, 1)) = (1, 2) imply ϕ((2, 0)) = (2b, 0). Now, consider the cycle C2 :(a + b +1, 0) ∼ (a +1, 1) ∼ (1, 2) ∼ (2b, 0) ∼ (2a + b, 1) ∼ (a + b +2, 2) ∼ (a + b +1, 0). So ϕ(C2):(b − a − 1, 0) ∼ (−a − b, 2) ∼ (−b, 1) ∼ ϕ((2b, 0)) ∼ ϕ((2a + b, 1)) ∼ ϕ((a + b +2, 2)) ∼ (b − a − 1, 0). Again ϕ((0, 0)) = (0, 0), ϕ((a +1, 1)) = (−a − b, 2) and ϕ((2b, 0)) ∼ (−b, 1) imply ϕ((2b, 0)) ∈{(−a+b, 2), (−2, 0)}. And ϕ(((a+b, 2)) = (a−1, 1), ϕ((a+1, 1)) = (−a−b, 2) and ϕ((a+b+2, 2)) ∼ (b−a−1, 0) imply ϕ((a+b+2, 2)) ∈{(−b−a+2, 2), (−1−2b+a, 1)}. ϕ((1, 2)) = (−b, 1) and ϕ((2a + b, 1)) ∼ ϕ((2b, 0)) imply

ϕ((2a + b, 1)) ∈{(−1+ a ± b, 0), (−b +2, 1), (−3b, 1), (−2a ± 1, 2). (16)

Case B(3)(a): If ϕ((a + b +2, 2)) = (−b − a +2, 2) then ϕ((a + b +1, 0)) = (b − a − 1, 0) and ϕ((2a + b, 1)) ∼ ϕ((a + b +2, 2)) imply

ϕ((2a + b, 1)) ∈{(a − 1+3b, 0), (−1 − b +2a ± 1, 1)}. (17)

From the Equations 16 and 17 we have, • either −1 − b +2a ± 1= −b + 2, which imply either 2a − 2b = 0 which is impossible by Lemma 4.1 or 4a − 2b = 0 which is possible only for n = 7 or 14. However, direct SageMath computation for n = 7 and 14 show that such ϕ does not exist. • or a − 1+3b = −1 − a ± b, i.e., 2a +4b = 0 which is possible only for n = 9 or 2a +2b = 0, which is impossible by Lemma 4.1. And ﬁnally direct SageMath computation for n = 9 show that such ϕ does not exist. • or −1 − b +2a ± 1 = −3b, i.e., 2a +2b =0or2a +2b − 2 = 0, both of which are impossible by Lemma 4.1. Hence ϕ((a + b +2, 2)) =(6 −b − a +2, 2). Case B(3)(b): If ϕ((a+b+2, 2)) = (a−1−2b, 1), then ϕ((a+b+1, 0)=(b−a−1, 0) and ϕ((2a + b, 1)) ∼ ϕ((a + b +2, 2)) imply

ϕ((2a + b, 1)) ∈{(b − a − 3, 0), (1 − b − 2a ± b, 2)}. (18)

From the Equations 16 and 18 we have,

12 • either b − a − 3= −1+ a ± b, i.e., 2a +2b =0or2a − 2b +2 = 0, both of which are impossible by Lemma 4.1.

• or 1 − b − 2a ± b = −2a ± 1. These gives rise two four equations, out of which three are impossible by Lemma 4.1, namely 2 = 0, 2b = 0 and 2a − 2b = 0. The only possibility which remains is 1 − b − 2a + b = −2a + 1 and it is an identity.

So assuming this to be the case, we have ϕ((2b, 0)) = (−2, 0), ϕ((a+b+2, 2)) = (a−1−2b, 1) and ϕ((2a + b, 1)) = (−2a +1, 2). Now consider the cycle C3 : (2b, 0) ∼ (2a + b, 1) ∼ (a + b +2, 2) ∼ (1 + a +3b, 0) ∼ (3a +1, 1) ∼ (3, 2) ∼ (2b, 0). So ϕ(C3):(−2, 0) ∼ (−2a +1, 2) ∼ (a − 1 − 2b, 1) ∼ ϕ((1+a+3b, 0)) ∼ ϕ((3a+1, 1)) ∼ ϕ((3, 2)) ∼ (−2, 0). Now, ϕ((2b, 0)) = (−2, 0), ϕ((2a+ b, 1)) = (−2a +1, 2) and ϕ((3, 2)) ∼ (−2, 0) imply ϕ((3, 2)) ∈ {(−3b, 1), (−2a − 1, 2)}. Again ϕ((2a + b, 1)) = (−2a +1, 2), ϕ((a + b +1, 0)) = (b − a − 1, 0) and ϕ((1 + a + 3b, 0)) ∼ (a − 1 − 2b, 1) imply ϕ((1 + a +3b, 0)) ∈{(1 − 2a − 2b, 2), (b − a − 3, 0)}. Finally ϕ((2b, 0)) = (−2, 0) and ϕ((3a +1, 1)) ∼ ϕ((3, 2)) imply

ϕ((3a +1, 1)) ∈{(−4, 0), (−3a ± b, 2), (−2 − 2b, 0), (−2b − a ± 1, 1)}. (19)

Case B(3)(b)(1): If ϕ((1+a+3b, 0)) = (1−2a−2b, 2) then ϕ((3a+1, 1)) ∼ ϕ((1+a+3b, 0)) implies ϕ((3a +1, 1)) ∈{(b − 2a − 2 ± b, 0), (a − 2 − 2b ± 1, 1)}. (20) From the Equations 19 and 20 we have,

• either b − 2a − 2 ± b = −4, i.e., 2a − 2b =0or2a − 2b − 2 = 0, both of which are impossible by Lemma 4.1.

• or b − 2a − 2 ± b = −2 − 2b, i.e., 2a − 2b =0or2a − 4b = 0, both of which are impossible by Lemma 4.1.

• or a − 2 − 2b ± 1 = −2b − a ± 1, i.e. 2a − 2b =0or2a =0or4a − 2b = 0. By Lemma 4.1, the ﬁrst two are impossible and the third one may hold only for n = 7 or 14. However, direct SageMath computation for n = 7 and 14 show that such ϕ does not exist.

So we have ϕ((1 + a +3b, 0)) =6 (1 − 2b − 2a, 2). Case B(3)(b)(2): If ϕ((1 + a +3b, 0)) = (b − a − 3, 0) then ϕ((3a +1, 1)) ∼ ϕ((1 + a +3b, 0)) implies

ϕ((3a +1, 1)) ∈{(a − 1 − 3b ± b, 1), (b − 1 − 3a ± 1, 2)}. (21)

From the Equations 19 and 21 we have,

• either a − 1 − 3b ± b = −2b − a ± 1, i.e., 2a =0or2a − 2b =0or2a − 2b − 2 = 0, all of which are impossible by Lemma 4.1.

13 • or 1 − b − 3a ± 1 = −3a ± b. Out of the four relations that we get, three of them (namely, 2 = 0, 2a − 2b =0 and 2b = 0) are invalid by Lemma 4.1 and the fourth is an identity, i.e., 1 − b − 3a − 1= −3a − b.

So we have ϕ((3a +1, 1)) = (−3a − b, 2), ϕ((3, 2)) = (−3b, 1), ϕ((a +1+3b, 0)) = (b − a−3, 0). Similarly we can show that ϕ((3a−1, 1)) = (−3a+b, 2) and ϕ((−a+1+3b, 0)) = (a + b − 3, 0). Now ϕ((2b, 0)) = (−2, 0), ϕ((3a +1, 1)) = (−3a − b, 2), ϕ((3a − 1, 1)) = (−3a + b, 2) and ϕ((4b, 0)) ∼ ϕ((3, 2)) = (−3b, 1) imply ϕ((4b, 0)) = (−4, 0). Proceeding in this way, we can show that ϕ((2kb, 0)) = (−2k, 0), for all k ∈ Z. So we have ϕ((2, 0)) = (−2a, 0), where k = a, which is a contradiction as we have shown earlier that ϕ((2, 0)) = (2b, 0) and 2b =6 −2a. Therefore ϕ((a + b +1, 0)) =(6 b − a − 1, 0). Case B(1): When ϕ((a+1, 1)) = (−a−b, 2), ϕ((1+a+b, 0)) = (2a+b, 1), ϕ((a+b, 2)) = (2b, 0) then we have the Equation 14. As 2a +2b + 2 = 0, i.e., a + b +1= −a − b − 1, then ϕ((−1−a−b, 0)) = (2a+b, 1). But from the Equations 9 and 10 we have ϕ((−1−a−b, 0)) =6 (2a + b, 1), which is a contradiction. Hence ϕ((1 + a + b, 0)) =6 (2a + b, 1). Case B(2): Now when ϕ((a +1, 1)) = (−a − b, 2), ϕ((a + b +1, 0)) = (a + b +1, 0), ϕ((a + b, 2)) = (a +1, 1) then we have the Equation 15. Consider the cycle C1 : (a + b + 1, 0) ∼ (a +1, 1) ∼ (1+2b, 2) ∼ (2a, 0) ∼ (b +2, 1) ∼ (a + b, 2) ∼ (a + b +1, 0). Then ϕ(C1):(a+b+1, 0) ∼ (−a−b, 2) ∼ ϕ((1+2b, 2)) ∼ ϕ((2a, 0)) ∼ ϕ((b+2, 1)) ∼ (a+1, 1) ∼ (a + b +1, 0). Now ϕ((a + b +1, 0)) = (a + b +1, 0)=(−a − b − 1, 0), ϕ((1, 2)) = (−b, 1) and ϕ((1 + 2b, 2)) ∼ (−a − b, 2) imply ϕ((1 + 2b, 2)) ∈ {(−1 − a + b, 0), (−b − 2, 1)}. Again, ϕ((b, 1)) = (1, 2), ϕ((a + b +1, 0)) = (a + b +1, 0) and ϕ((b +2, 1)) ∼ ϕ((a +1, 1)) imply ϕ((b +2, 1)) ∈ {(b + a − 1, 0), (1+ 2b, 2)}. Also ϕ((a +1, 1)) = (−a − b, 2) and ϕ((2a, 0)) ∼ ϕ((1+2b, 2)) imply

ϕ((2a, 0)) ∈{(−b − 1+ a ± b, 1), (−a − b +2, 2), (−1 − 2a ± 1, 0), (−a − 3b, 2)}. (22)

Case B(2)(a): Now if ϕ((b +2, 1)) = (b + a − 1, 0) then ϕ((a + b, 2)) = (a +1, 1) and ϕ((2a, 0)) ∼ ϕ((b +2, 1)) imply

ϕ((2a, 0)) ∈{(a +1 − 2b, 1), (1 + b − a ± 1, 2)}. (23)

From the Equations 22 and 23, we have

• either 1 + b − a ± 1 = −a − b + 2, i.e., 2b =0or2a − 2b = 0, both of which are impossible by Lemma 4.1.

• or 1+ b − a ± 1= −a − 3b, i.e., 4b = 0 which is impossible or 4a +2b = 0, which, by Lemma 4.1, holds only if n = 18. However, direct SageMath computation for n = 18 shows that such ϕ does not exist.

• or a +1 − 2b = −b − 1+ a ± b, i.e., 2 = 0 or 2a − 2b = 0, both of which are impossible by Lemma 4.1.

14 Hence ϕ((b +2, 1)) =(6 b + a − 1, 0). Case B(2)(b): Now if ϕ((b +2, 1)) = (1+2b, 2) then ϕ((a + b, 2)) = (a +1, 1) and ϕ((2a, 0)) ∼ ϕ((b +2, 1)) imply

ϕ((2a, 0)) ∈{(a +3, 1), (b +2a ± b, 0)}. (24)

From the Equations 22 and 24, we have

• either a +3= −b − 1+ a ± b, i.e., 4 = 0 or 2a +4b = 0, which, by Lemma 4.1, can hold only if n = 9. However, direct SageMath computation for n = 9 shows that such ϕ does not exist.

• or b +2a ± b = −1 − 2a ± 1, i.e., 2(a + b +2) = 0 or 4a = 0or2a +4b = 0 or 4a +2b = 0. By Lemma 4.1, the ﬁrst two are impossible and the next two can hold only if n = 9 or 18. But those are also ruled out by SageMath computation.

Hence ϕ((b +2, 1)) =6 (1+2b, 2) and hence ϕ((a +1, 1)) =6 (−a − b, 2), i.e., Case B can not hold. Case C: If ϕ((a +1, 1)) = (−a + b, 2) then ϕ((1, 2)) = (−b, 1) and ϕ((a + b +1, 0)) ∼ ϕ((a +1, 1)) imply

ϕ((1 + a + b, 0)) ∈{(−b +2, 1), (−1+ a ± b, 0)} (25)

From the Equations 8 and 25, we have

• either −b +2=2a ± b, i.e., 2a − 2b = 0 or 2(a + b − 1) = 0, both of which are impossible by Lemma 4.1.

• or −1+ a ± b = b − a ± 1, i.e., 2a =0or2a − 2b =0 or 2(a − b − 1) =0 and all of them are ruled out by Lemma 4.1.

• or −1+ a ± b = b + a ± 1. This gives rise two four conditions, out of which three (namely, 2 = 0, 2a +2b =0 and 2b = 0) are ruled out by Lemma 4.1 and the fourth one is the identity −1+ a + b = b + a − 1.

So we have ϕ((a + b +1, 0)) = (a − 1+ b, 0), ϕ(((a + b, 2)) = (a +1, 1) and ϕ((a +1, 1)) = (−a + b, 2). Similarly we can show that ϕ((−a+b+1, 0)) = (−a−1+b, 0), ϕ(((a−b, 2)) = (a−1, 1) and ϕ((a − 1, 1)) = (−a − b, 2). Now ϕ((a+b, 2)) = (a+1, 1), ϕ(((a−b, 2)) = (a−1, 1) and ϕ((2, 0)) ∼ ϕ((b, 1)) = (1, 2) imply ϕ((2, 0)) = (2b, 0). Consider the cycle C2 : (a + b +1, 0) ∼ (a +1, 1) ∼ (1, 2) ∼ (2b, 0) ∼ (2a + b, 1) ∼ (a + b + 2, 2) ∼ (a + b + 1, 0). So ϕ(C2):(a − 1 + b, 0) ∼ (−a + b, 2) ∼ (−b, 1) ∼ ϕ((2b, 0)) ∼ ϕ((2a + b, 1)) ∼ ϕ((a + b +2, 2)) ∼ (a − 1+ b, 0). Now, ϕ((0, 0)) = (0, 0), ϕ((a+1, 0)) = (−a+b, 2) and ϕ((2b, 0)) ∼ (−b, 1) imply ϕ((2b, 0)) ∈{(−a−b, 2), (−2, 0)}. Again, ϕ(((a+b, 2)) = (a+1, 1), ϕ((a+1, 1)) = (−a+b, 2) and ϕ((a+b+2, 2)) ∼ (a−1+b, 0)

15 imply ϕ((a + b +2, 2)) ∈ {(b − a +2, 2), (1 − 2b + a, 1)}. Also, ϕ((1, 2)) = (−b, 1) and ϕ((2a + b, 1)) ∼ ϕ((2b, 0)) imply

ϕ((2a + b, 1)) ∈{(−1 − a ± b, 0), (−b − 2, 1), (−3b, 1), (−2a ± 1, 2)}. (26)

Case C(1): If ϕ((a + b +2, 2)) = (b − a +2, 2) then ϕ((a + b +1, 0)) = (a + b − 1, 0) and ϕ((2a + b, 1)) ∼ ϕ((a + b +2, 2)) imply

ϕ((2a + b, 1)) ∈{(a − 1+3b, 0), (1 − b +2a ± 1, 1)}. (27)

From the Equations 26 and 27 we have • either 1 − b +2a ± 1= −b − 2, i.e., 2a +2b =0or4a +2b = 0. By Lemma 4.1, the ﬁrst is an impossibility and the second one can hold only if n = 18. However, that is also ruled out by SageMath computation for n = 18.

• or a − 1+3b = −1 − a ± b, i.e., 2a +2b =0or2a +4b = 0. By Lemma 4.1, the ﬁrst is an impossibility and the second one can hold only if n = 9. However, that is also ruled out by SageMath computation for n = 9.

• or 1 − b +2a ± 1= −3b, i.e., 2a +2b = 0, which is impossible by the Lemma 4.1 but,

2a +2b +2=0 may hold. (28)

When ϕ((a + b +1, 0)) = (a − 1+ b, 0), ϕ(((a + b, 2)) = (a +1, 1) and ϕ((a +1, 1)) = (−a + b, 2) then we have the Equation 28. As 2a +2b + 2 = 0, i.e., a + b +1= −a − b − 1, then ϕ((−1 − a − b, 0)) = (a − 1+ b, 0). But from the Equations 9 and 10, we have ϕ((−1 − a − b, 0)) =6 (a − 1+ b, 0), which is a contradiction. Hence ϕ((a + b +2, 2)) =6 (b − a +2, 2). Case C(2): If ϕ((a + b +2, 2)) = (a +1 − 2b, 1), then ϕ((a + b +1, 0)=(a − 1+ b, 0) and ϕ((2a + b, 1)) ∼ ϕ((a + b +2, 2)) imply

ϕ((2a + b, 1)) ∈{(b + a − 3, 0), (1+ b − 2a ± b, 2)}. (29)

From the Equations 26 and 29 we have, • either b + a − 3= −1 − a ± b, i.e., 2a − 2b =0or2a +2b − 2 = 0, both of which are impossible by Lemma 4.1.

• or 1+ b − 2a ± b = −2a ± 1, This gives rise to four conditions, out of which three (namely 2b =0, 2 − 0 and 2a +2b = 0) are ruled out by Lemma 4.1 and the fourth one is the identity 1 + b − 2a − b = −2a + 1. So we have ϕ((2a + b, 1)) = (−2a +1, 2). Also previously, we had ϕ((2b, 0)) = (−2, 0) and ϕ((a + b +2, 2)) = (1 − 2b + a, 1). Now consider the cycle C3 : (2b, 0) ∼ (2a + b, 1) ∼ (a + b +2, 2) ∼ (1 + a +3b, 0) ∼ (3a +1, 1) ∼ (3, 2) ∼ (2b, 0). So ϕ(C3):(−2, 0) ∼ (−2a +1, 2) ∼ (1 + a − 2b, 1) ∼

16 ϕ((1+a+3b, 0)) ∼ ϕ((3a+1, 1)) ∼ ϕ((3, 2)) ∼ (−2, 0). Now, ϕ((2b, 0)) = (−2, 0), ϕ((2a+ b, 1)) = (−2a +1, 2) and ϕ((3, 2)) ∼ (−2, 0) imply ϕ((3, 2)) ∈ {(−3b, 1), (−2a − 1, 2)}. Again ϕ((2a + b, 1)) = (−2a +1, 2), ϕ((a + b +1, 0)) = (a + b − 1, 0) and ϕ((1 + a + 3b, 0)) ∼ (1 + a − 2b, 1) imply ϕ((1 + a +3b, 0)) ∈{(2b +1 − 2a, 2), (a + b − 3, 0)}. Also, ϕ((2b, 0)) = (−2, 0) and ϕ((3a +1, 1)) ∼ ϕ((3, 2)) imply

ϕ((3a +1, 1)) ∈{(−4, 0), (−3a ± b, 2), (−2 − 2b, 0), (−2b − a ± 1, 1)}. (30)

Case C(2)(a): If ϕ((1 + a +3b, 0)) = (2b +1 − 2a, 2), then ϕ((3a +1, 1)) ∼ ϕ((1 + a +3b, 0)) implies

ϕ((3a +1, 1)) ∈{(2a + b − 2 ± b, 0), (2 + a − 2b ± 1, 1)}. (31)

From the Equations 30 and 31 we have

• either 2 + a − 2b ± 1= −2b − a ± 1, i.e. 2a +2b =0or2a = 0 (which are impossible by Lemma 4.1) or 4a +2b = 0 which can hold only if n = 18. But direct SageMath computation for n = 18 ruled out this case.

• or 2a + b − 2 ± b = −2 − 2b, i.e., 2a + 2b = 0 (impossible by Lemma 4.1) or 2a +4b = 0, which can hold only if n = 9. But direct SageMath computation ruled out this possibility.

• or 2a + b − 2 ± b = −4, i.e., 2a +2b = 0, which is impossible by the Lemma 4.1 but

2a +2b +2=0 may hold. (32)

Thus, if ϕ((a + b +1, 0)) = (a − 1+ b, 0), ϕ(((a + b, 2)) = (a +1, 1) and ϕ((a +1, 1)) = (−a+b, 2) holds, then we have 2a+2b+2=0. As2a+2b+2 = 0, i.e., a+b+1= −a−b−1, then ϕ((−1 − a − b, 0)) = (a − 1 + b, 0). But from the Equations 9 and 10 we have ϕ((−1 − a − b, 0)) =(6 a − 1+ b, 0), which is a contradiction. Thus, Equation 32 does not hold. So we have ϕ((1 + a +3b, 0)) =6 (2b +1 − 2a, 2). Case C(2)(b): If ϕ((1+a+3b, 0)) = (a+b−3, 0), then ϕ((3a+1, 1)) ∼ ϕ((1+a+3b, 0)) implies ϕ((3a +1, 1)) ∈{(1 + a − 3b ± b, 1), (b +1 − 3a ± 1, 2)}. (33) From the Equations 30 and 33 we have,

• either 1 + a − 3b ± b = −2b − a ± 1, i.e., 2a =0or2a +2b =0or2a − 2b = 0 or 2a − 2b + 2 = 0, all of which are impossible by Lemma 4.1.

• or b +1 − 3a ± 1= −3a ± b. This gives rise to four conditions. Out of which three (namely, 2 = 0, 2a +2b =0 and 2b = 0) are ruled out by Lemma 4.1 and fourth one is the identity b +1 − 3a − 1= −3a + b.

17 So we have ϕ((3a+1, 1)) = (−3a+b, 2), ϕ((3, 2)) = (−3b, 1), ϕ((a+1+3b, 0)) = (a+b−3, 0). Similarly we can show that ϕ((3a − 1, 1)) = (−3a − b, 2) and ϕ((−a +1+3b, 0)) = (−a + b − 3, 0). Now ϕ((2b, 0)) = (−2, 0), ϕ((3a +1, 1)) = (−3a + b, 2), ϕ((3a − 1, 1)) = (−3a − b, 2) and ϕ((4b, 0)) ∼ ϕ((3, 2)) = (−3b, 1) imply ϕ((4b, 0)) = (−4, 0). Proceeding this way, we can show that ϕ((2kb, 0)) = (−2k, 0), for all k ∈ Z. So we have ϕ((2, 0)) = (−2a, 0), where k = a, which is a contradiction as we have shown that ϕ((2, 0)) = (2b, 0) and 2b =6 −2a. Therefore we have ϕ((a +1, 1)) =(6 −a + b, 2) and Case C can not hold. As none of the Cases A, B and C hold, the assumption that ϕ((1, 2)) = (−b, 1) is wrong. Hence the lemma follows. Theorem 4.3. If ϕ ∈ G and ϕ((0, 0)) = (0, 0), ϕ((b, 1)) = (1, 2), ϕ((1, 2)) = (b, 1) then n =7 or 14. Proof: Consider the cycle C0 : (0, 0) ∼ (b, 1) ∼ (a + b, 2) ∼ (a + b +1, 0) ∼ (a +1, 1) ∼ (1, 2) ∼ (0, 0). Then ϕ(C0) : (0, 0) ∼ (1, 2) ∼ ϕ((a + b, 2)) ∼ ϕ((a + b +1, 0)) ∼ ϕ((a + 1, 1)) ∼ (b, 1) ∼ (0, 0)}. ϕ((0, 0)) = (0, 0) and ϕ((a + b, 2)) ∼ (1, 2) imply ϕ((a + b, 2)) ∈ {(2b, 0), (a ± 1, 1). ϕ((0, 0)) = (0, 0) and ϕ((a + 1, 1)) ∼ (b, 1) imply ϕ((a + 1, 1)) ∈ {(2, 0), (a ± b, 2)}. Now ϕ((b, 1)) = (1, 2) and ϕ((a + b +1, 0)) ∼ ϕ((a + b, 2)) imply ϕ((1+a+b, 0)) ∈{(2a±b, 1), (3, 2), (1+2b, 2), (b+a±1, 0), (1−2b, 2), (b−a±1, 0)}. (34) Depending upon the value of ϕ((a +1, 1)), we split it into three cases a, b and c. Case a: If ϕ((a +1, 1)) = (2, 0) then ϕ((1 + a + b, 0)) ∼ ϕ((a +1, 1)) and ϕ((1, 2)) = (−b, 1)) imply ϕ((1 + a + b, 0)) ∈{(3b, 1), (2a ± 1, 2)} (35) From the Equation 34 and 35 we have, • either 3b =2a ± b, i.e., 2a − 2b =0or4b − 2a = 0, which are impossible by Lemma 4.1. • or 2a ± 1=1+2b, i.e, 2a − 2b =0or2a − 2b − 2 = 0, which are impossible by Lemma 4.1. • or 2a±1=1−2b, i.e, 2a+2b =0or2a+2b−2 = 0, which are impossible by Lemma 4.1. • or 2a ± 1 = 3, i.e., 2a = 2, i.e., 2a − 2b =0or2a = 4, i.e., 4a − 2b = 0. By Lemma 4.1, 2a − 2b = 0 can not hold and 4a − 2b = 0 can hold only if n = 7, 14. However direct SageMath computation for n =7, 14 shows that such ϕ does not exist. Therefore we have ϕ((a +1, 1)) =6 (2, 0) and hence Case a can not hold. Case b: If ϕ((a+1, 1)) = (a+b, 2) then ϕ((1+a+b, 0)) ∼ ϕ((a+1, 1)) and ϕ((1, 2)) = (b, 1) imply ϕ((1 + a + b, 0)) ∈{(b +2, 1), (1 + a ± b, 0)} (36) From the Equation 34 and 36 we have,

18 • either b +2=2a ± b, i.e., 2a = 2, i.e., 2a − 2b =0or2a − 2b − 2 = 0, which are impossible. • or 1+ a ± b = b − a ± 1, i.e., 2a =0or2a + 2 = 0, i.e., 2a +2b =0or2a − 2b =0 or 2a − 2b + 2 = 0, which are impossible. • or 1+a±b = b+a±1. This gives rise four equations out of which three are impossible by Lemma 4.1, namely 2 = 0, 2b =0,2b−2 = 0, i.e., 2a−2b = 0. The only possiblity which remains is 1 + a + b = b + a + 1, which is an identity. So asssuming this to be the case we have ϕ((a +1, 1)) = (a + b, 2), ϕ((a + b +1, 0)) = (a + b +1, 0) and ϕ((a + b, 2)) = (a +1, 1). Case c: If ϕ((a+1, 1)) = (a−b, 2) then ϕ((1+a+b, 0)) ∼ ϕ((a+1, 1)) and ϕ((1, 2)) = (b, 1) imply ϕ((1 + a + b, 0)) ∈{(b − 2, 1), (1 − a ± b, 0)} (37) From the Equation 34 and 37 we have, • either b − 2=2a ± b, i.e., 2a + 2 = 0, i.e., 2a +2b =0or2a − 2b + 2 = 0, which are impossible. • or 1 − a ± b = b + a ± 1, i.e., 2 = 0 or 2a − 2 = 0, i.e., 2a − 2b =0or2a +2b =0 or 2a +2b − 2 = 0, which are impossible. • or 1−a±b = b−a±1,This gives rise four equations out of which three are impossible by Lemma 4.1, namely 2 = 0, 2b =0,2b−2 = 0, i.e., 2a−2b = 0. The only possiblity which remains is 1 − a + b = b − a + 1, which is an identity. So asssuming this to be the case we have ϕ((a +1, 1)) = (a − b, 2), ϕ((a + b +1, 0)) = (b − a +1, 0) and ϕ((a + b, 2)) = (a − 1, 1). Combining the feasible cases in Case b and Case b, we have ϕ((a +1, 1)) ∈{(a ± b, 2)}, ϕ((a + b +1, 0)) ∈{(b ± a +1, 0)} and ϕ((a + b, 2)) ∈{(a ± 1, 1)}. ′ Now consider the cycle C0 : (0, 0) ∼ (1, 2) ∼ (a − b, 2) ∼ (b − a +1, 0) ∼ (a − 1, 1) ∼ ′ (1, 2) ∼ (0, 0). Then ϕ(C0) : (0, 0) ∼ (1, 2) ∼ ϕ((a − b, 2)) ∼ ϕ((b − a +1, 0)) ∼ ϕ((a − 1, 1)) ∼ (b, 1) ∼ (0, 0)}. ϕ((0, 0)) = (0, 0) and ϕ((a − b, 2)) ∼ (1, 2) imply ϕ((a − b, 2)) ∈ {(2b, 0), (a ± 1, 1). ϕ((0, 0)) = (0, 0) and ϕ((a − 1, 1)) ∼ (b, 1) imply ϕ((a − 1, 1)) ∈ {(2, 0), (a ± b, 2)}. Now ϕ((b, 1)) = (1, 2) and ϕ((b − a +1, 0)) ∼ ϕ((a − b, 2)) imply

ϕ((b−a+1, 0)) ∈{(2a±b, 1), (3, 2), (1+2b, 2), (b+a±1, 0), (1−2b, 2), (b−a±1, 0)}. (38)

Case a′: If ϕ((a−1, 1)) = (2, 0) then ϕ((b−a+1, 0)) ∼ ϕ((a−1, 1)) and ϕ((1, 2)) = (−b, 1)) imply ϕ((b − a +1, 0)) ∈{(3b, 1), (2a ± 1, 2)} (39) From the Equation 38 and 39 we have, • either 3b = 2a ± b, i.e., 2a − 2b =0or4b − 2a = 0, which are impossible by the Lemma 4.1.

19 • or 2a ± 1=1+2b, i.e, 2a − 2b =0or2a − 2b − 2 = 0, which are impossible by the Lemma 4.1.

• or 2a ± 1=1 − 2b, i.e, 2a +2b =0or2a +2b − 2 = 0, which are impossible by the Lemma 4.1.

• or 2a ± 1 = 3, i.e., 2a = 2, i.e., 2a − 2b =0or2a = 4, i.e., 4a − 2b = 0. By Lemma 4.1 2a − 2b = 0 can not hold and 4a − 2b = 0 can hold only if n = 7, 14. Therefore we have ϕ((a − 1, 1)) = (2, 0) only if n =7, 14.

Case b′: If ϕ((a − 1, 1)) = (a + b, 2) then ϕ((b − a + 1, 0)) ∼ ϕ((a − 1, 1)) and ϕ((1, 2)) = (b, 1) imply

ϕ((b − a +1, 0)) ∈{(b +2, 1), (1+ a ± b, 0)} (40)

From the Equation 38 and 40 we have,

• either b +2=2a ± b, i.e., 2a = 2, i.e., 2a − 2b =0or2a − 2b − 2 = 0, which are impossible.

• or 1+ a ± b = b − a ± 1, i.e., 2a =0or2a + 2 = 0, i.e., 2a +2b =0or2a − 2b =0 or 2a − 2b + 2 = 0, which are impossible.

• or 1+a±b = b+a±1. This gives rise four equations out of which three are impossible by Lemma 4.1, namely 2 = 0, 2b =0,2b−2 = 0, i.e., 2a−2b = 0. The only possiblity which remains is 1 + a + b = b + a + 1, which is an identity. So asssuming this to be the case we have ϕ((a − 1, 1)) = (a + b, 2), ϕ((b − a +1, 0)) = (a + b +1, 0) and ϕ((a − b, 2)) = (a +1, 1).

Case c′: If ϕ((a − 1, 1)) = (a − b, 2) then ϕ((b − a + 1, 0)) ∼ ϕ((a − 1, 1)) and ϕ((1, 2)) = (b, 1) imply

ϕ((b − a +1, 0)) ∈{(b − 2, 1), (1 − a ± b, 0)} (41)

From the Equation 38 and 41 we have,

• either b − 2=2a ± b, i.e., 2a + 2 = 0, i.e., 2a +2b =0or2a − 2b + 2 = 0, which are impossible.

• or 1 − a ± b = b + a ± 1, i.e., 2 = 0 or 2a − 2 = 0, i.e., 2a − 2b =0or2a +2b =0 or 2a +2b − 2 = 0, which are impossible.

• or 1−a±b = b−a±1, this gives rise four equations out of which three are impossible by Lemma 4.1, namely 2 = 0, 2b =0,2b−2 = 0, i.e., 2a−2b = 0. The only possiblity which remains is 1 − a + b = b − a + 1, which is an identity. So asssuming this to be the case we have ϕ((a − 1, 1)) = (a − b, 2), ϕ((b − a +1, 0)) = (b − a +1, 0) and ϕ((a − b, 2)) = (a − 1, 1).

20 Combining these three cases we have ϕ((a − 1, 1)) ∈ {(a ± b, 2)}, ϕ((b − a + 1, 0)) ∈ {(b ± a +1, 0)}, ϕ((a − b, 2)) ∈{(a ± 1, 1)}. ϕ((a − 1, 1)) = (2, 0), ϕ((b − a +1, 0)) = (3, 2), ϕ((a − b, 2)) = (2b, 0) only if n =7, 14. Depending upon the values of ϕ((a+1, 1)) and ϕ((a−1, 1)) (that we get from Cases a, b, c and a′, b′, c′), we have four diﬀerent cases namely Case A: ϕ((a+1, 1)) = (a+b, 2) and ϕ((a−1, 1)) = (2, 0) (only if n =7, 14), Case B: ϕ((a+1, 1)) = (a−b, 2) and ϕ((a−1, 1)) = (2, 0) (only if n =7, 14), Case C: ϕ((a +1, 1)) = (a − b, 2) and ϕ((a − 1, 1)) = (a + b, 2) and Case D: ϕ((a +1, 1)) = (a + b, 2) and ϕ((a − 1, 1)) = (a − b, 2). However, before resolving these four cases, we prove a claim which will be crucial in the following proof. Claim: ϕ((−1 − a − b, 0)) ∈{(−1+ a ± b, 0), (−b ± 2, 1), (−1 − a ± b, 0), (−2a ± 1, 2), (−3b, 1), (−b + a ± 1, 0, 0), (−1 ± 2b, 2), (−b − a ± 1, 0), (−2a ± b, 1), (−3, 2)}. Proof of Claim: ϕ((0, 0)) = (0, 0), ϕ((b, 1)) = (1, 2), ϕ((1, 2)) = (b, 1), ϕ((−b, 1)) ∼ ϕ((0, 0)) = (0, 0) and imply ϕ((−1, 2)) ∼ ϕ((0, 0)) = (0, 0) imply ϕ((−b, 1)), ϕ((−1, 2)) ∈ {(−b, 1), (−1, 2)}. ′′ Case 1: Let ϕ((−b, 1)) = (−b, 1) and ϕ((−1, 2)) = (−1, 2). Consider the cycle C0 : (0, 0) ∼ (−b, 1) ∼ (−a + −, 2) ∼ (−a − b − 1, 0) ∼ (−a − 1, 1) ∼ (−1, 2) ∼ (0, 0). Then ′′ ϕ(C0 ) : (0, 0) ∼ (−b, 1) ∼ ϕ((−a − b, 2)) ∼ ϕ((−a − b − 1, 0)) ∼ ϕ((−a − 1, 1)) ∼ (−1, 2) ∼ (0, 0)}. ϕ((−a − b, 2)) ∼ (−b, 1) and ϕ((0, 0)) = (0, 0) imply ϕ((−a − b, 2)) ∈ {(−a±b, 2), (−2, 0)}. ϕ((−a−1, 1)) ∼ (−1, 2) and ϕ((0, 0)) = (0, 0) imply ϕ((−a−1, 1)) ∈ {(−a ± 1, 1), (−2b, 0)}. ϕ((−a − b − 1, 0)) ∼ ϕ((−a − b, 2)) and ϕ((−b, 1)) = (−b, 1) imply ϕ((−a−b−1, 0)) ∈{(−1+a±b, 0), (−b+2, 1), (−1−a±b, 0), (−b−2, 1), (−2a±1, 2), (−3b, 1)}. (42) Case 2: Let ϕ((−b, 1)) = (−1, 2) and ϕ((−1, 2)) = (−b, 1). Again consider the cycle ′′ C0 : (0, 0) ∼ (−b, 1) ∼ (−a + −, 2) ∼ (−a − b − 1, 0) ∼ (−a − 1, 1) ∼ (−1, 2) ∼ (0, 0). ′′ Then ϕ(C0 ) : (0, 0) ∼ (−1, 2) ∼ ϕ((−a − b, 2)) ∼ ϕ((−a − b − 1, 0)) ∼ ϕ((−a − 1, 1)) ∼ (−1, 2) ∼ (0, 0)}. ϕ((−a − b, 2)) ∼ (−1, 2) and ϕ((0, 0)) = (0, 0) imply ϕ((−a − b, 2)) ∈ {(−a±1, 1), (−2b, 0)}. ϕ((−a−1, 1)) ∼ (−b, 1) and ϕ((0, 0)) = (0, 0) imply ϕ((−a−1, 1)) ∈ {(−a ± b, 2), (−2, 0)}. ϕ((−a − b − 1, 0)) ∼ ϕ((−a − b, 2)) and ϕ((−b, 1)) = (−1, 2) imply ϕ((−a−b−1, 0)) ∈{(−b+a±1, 0), (−1+2b, 2), (−b−a±1, 0), (−1−2b, 2), (−2a±b, 1), (−3, 2)}. (43) Case A: Let ϕ((a +1, 1)) = (a + b, 2), ϕ((a + b +1, 0)) = (b + a +1, 0), ϕ((a + b, 2)) = (a +1, 1), ϕ((a − 1, 1)) = (2, 0), ϕ((b − a +1, 0)) = (3, 2) and ϕ((a − b, 2)) = (2b, 0). This map exists only if n =7, 14. This can also be checked with direct Sagemath computation. Case B: Let ϕ((a +1, 1)) = (a − b, 2), ϕ((a + b +1, 0)) = (b − a +1, 0), ϕ((a + b, 2)) = (a − 1, 1), ϕ((a − 1, 1)) = (2, 0), ϕ((b − a +1, 0)) = (3, 2) and ϕ((a − b, 2)) = (2b, 0). Such ϕ can hold only if n = 7, 14. However direct SageMath computation for n = 7, 14 shows that such ϕ does not exist, hence this map does not exist for all n. Case C: Let ϕ((a +1, 1)) = (a − b, 2), ϕ((a + b +1, 0)) = (b − a +1, 0), ϕ((a + b, 2)) = (a−1, 1), ϕ((a−1, 1)) = (a+b, 2), ϕ((b−a+1, 0)) = (b+a+1, 0), ϕ((a−b, 2)) = (a+1, 1). Now ϕ((a + b, 2)) = (a − 1, 1), ϕ((a − b, 2)) = (a +1, 1) and ϕ((2, 0)) ∼ ϕ((b, 1)) = (1, 2) imply ϕ((2, 0)) = (2b, 0).

21 Consider the cycle C1 : (2, 0) ∼ (b, 1) ∼ (a + b, 2) ∼ (a + b +1, 0) ∼ (2b + a +1, 1) ∼ (2a +1, 2) ∼ (2, 0). then ϕ(C1):(2b, 0) ∼ (1, 2) ∼ (a − 1, 1) ∼ (b − a +1, 0) ∼ ϕ((2b + a + 1, 1)) ∼ ϕ((2a +1, 2)) ∼ (2b, 0). ϕ((2b + a +1, 1)) ∼ (b − a +1, 0), ϕ((a +1, 1)) = (a − b, 2) and ϕ((a + b, 2)) = (a − 1, 1) imply ϕ((2b + a +1, 1)) ∈ {(a − 1+2b, 1), (2 − b + a, 2)}. Now ϕ((2a +1, 2)) ∼ (2b, 0) and ϕ((b, 1)) = (1, 2) imply

ϕ((2a +1, 2)) ∈{(2a ± b, 1), (3, 2)}. (44)

Case C(a): If ϕ((2b + a +1, 1)) = (a − 1+2b, 1) then ϕ((2b + a +1, 1)) ∼ ϕ((2a +1, 2)) and ϕ((a + b +1, 0)) = (b − a +1, 0) imply

ϕ((2a +1, 2)) ∈{(1 − b +2a ± b, 2), (b − a +3, 0)}. (45)

From the Equation 44 and 45 we have, 1−b+2a±b = 3, i.e., 2a−2b−2=0or2a−2 = 0, i.e., 2a−2b = 0, which are impossible by the Lemma 4.1, hence ϕ((2b+a+1, 1)) =(6 a−1+2b, 1). Case C(b): If ϕ((2b + a +1, 1)) = (2 − b + a, 2) then ϕ((2b + a +1, 1)) ∼ ϕ((2a +1, 2)) and ϕ((a + b +1, 0)) = (b − a +1, 0) imply

ϕ((2a +1, 2)) ∈{(2a − 1+ b ± 1, 1), (3b − a +1, 0)}. (46)

From the Equation 44 and 46 we have, 2a−1+b±1=2a±b, This gives rise four equations out of which three are impossible by Lemma 4.1, namely 2 = 0, 2b = 0, 2b − 2 = 0, i.e., 2a − 2b = 0. The only possiblity which remains is 2a + b = 2a + b, which is an identity. So asssuming this to be the case we have ϕ((2a +1, 2)) = (2a + b, 1), ϕ((2b + a +1, 1)) = (2 − b + a, 2). ′ Now consider the cycle C1 : (2, 0) ∼ (b, 1) ∼ (a+b, 2) ∼ (1+a−b, 0) ∼ (2b+1−a, 1) ∼ ′ (2a − 1, 2) ∼ (2, 0). Then ϕ(C1) : (2b, 0) ∼ (1, 2) ∼ (a − 1, 1) ∼ ϕ((1 + a − b, 0)) ∼ ϕ((2b +1 − a, 1)) ∼ ϕ((2a − 1, 2)) ∼ (2b, 0). ϕ((2a − 1, 2)) ∼ (2b, 0), ϕ((b, 1)) = (1, 2) and ϕ((2a +1, 2)) = (2a + b, 1) imply ϕ((2a − 1, 2)) ∈ {(2a − b, 1), (3, 2)}. Now ϕ((1 + a − b, 0)) ∼ (a − 1, 1), ϕ((a + b + 1, 0)) = (b − a + 1, 0) and ϕ((b, 1)) = (1, 2) imply ϕ((1 + a − b, 0)) ∈ {(b − a − 1, 0), (1 − 2b, 2)}. ϕ((2b +1 − a, 1)) ∼ ϕ((1 + a − b, 0)) and ϕ((a + b, 2)) = (a − 1, 1) imply

ϕ((2b +1 − a, 1)) ∈{(a − 1 − 2b, 1), (1 − b − a ± 1, 2), (a − 3, 1), (b − 2a ± b, 0)}. (47)

Case C(b)(1): If ϕ((2a − 1, 2)) = (3, 2) then ϕ((2b +1 − a, 1)) ∼ ϕ((2a − 1, 2)) and ϕ((2, 0)) = (2b, 0) imply

ϕ((2b +1 − a, 1)) ∈{(4b, 0), (3a ± 1, 1)}. (48)

From the Equations 47 and 48 we have,

• either 4b = b − 2a ± b, i.e., 2a +2b =0or2a +4b = 0, which are impossible by the Lemma 4.1.

22 • 3a ± 1= a − 3, i.e., 2a + 4 = 0, i.e., 4a +2b =0or2a + 2 = 0, i.e., 2a +2b = 0. By Lemma 4.1 2a +2b = 0 is impossible and 4a +2b = 0 holds only if n = 18. But this is also ruled out by SageMath computation for n = 18.

• or 3a ± 1= a − 1 − 2b, i.e., 2a +2b = 0, which is impossible by the Lemma 4.1, but 2a+2b+2 = 0 may hold. We have this Equation when ϕ((a+b+1, 0)) = (b−a+1, 0). So 2a+2b+2 = 0 imply a+b+1 = −a−b−1, hence ϕ((−a−b−1, 0)) = (b−a+1, 0). But from the Equations 42 and 43 we have ϕ((−a − b − 1, 0)) =(6 b − a +1, 0), which is a contradiction. Hence ϕ((2a − 1, 2)) =6 (3, 2).

Case C(b)(2): If ϕ((2a − 1, 2)) = (2a − b, 1) then ϕ((2b +1 − a, 1)) ∼ ϕ((2a − 1, 2)) and ϕ((2, 0)) = (2b, 0) imply

ϕ((2b +1 − a, 1)) ∈{(2b − 2, 0), (2 − a ± b, 2)}. (49)

From the Equations 47 and 49 we have,

• either 2b − 2= b − 2a ± b, i.e., 2a − 2 = 0, i.e., 2a − 2b =0or2b +2a − 2 = 0, which are impossible by the Lemma 4.1.

• or 2 − a ± b = 1 − b − a ± 1, This gives rise four equations out of which three are impossible by Lemma 4.1, namely 2 = 0, 2b =0,2b+2 = 0, i.e., 2a+2b = 0. The only possiblity which remains is 2 − a − b =2 − b − a, which is an identity. So asssuming this to be the case we have ϕ((2b+1−a, 0)) = (2−b−a, 0), ϕ((2a−1, 2)) = (2a−b, 1) and ϕ((1 + a − b, 0)) = (b − a − 1, 0).

Now ϕ((3b, 1)) ∼ ϕ((2, 0)) = (2b, 0), ϕ((2a +1, 2)) = (2a + b, 1) and ϕ((2a − 1, 2)) = (2a − b, 1) imply ϕ((3b, 1)) = (3, 2). Consider the cycle C2 : (2, 0) ∼ (3b, 1) ∼ (3a + b, 2) ∼ (3+ a + b, 0) ∼ (2b + a +1, 1) ∼ (2a +1, 2) ∼ (2, 0). Then ϕ(C2) : (2b, 0) ∼ (3, 2) ∼ ϕ((3a + b, 2) ∼ ϕ((3 + a + b, 0)) ∼ (a − b +2, 2) ∼ (2a + b, 1) ∼ (2b, 0). ϕ((3a + b, 2)) ∼ (3, 2) and ϕ((2, 0)) = (2b, 0) imply ϕ((3a + b, 2)) ∈{(4b, 0), (3a ± 1, 1)}. ϕ((3 + a + b, 0)) ∼ (a − b +2, 2), ϕ((a + b +1, 0)) = (b − a +1, 0) and ϕ((2a +1, 2)) = (2a + b, 1) imply

ϕ((3 + a + b, 0)) ∈{(2a + b − 2, 1), (3b − a +1, 0)}. (50)

Case C(b)(2)(a): If ϕ((3a + b, 2)) = (4b, 0) then ϕ((3 + a + b, 0)) ∼ ϕ((3a + b, 2)) and ϕ((3b, 1)) = (3, 2) imply

ϕ((3 + a + b, 0)) ∈{(4a ± b, 1), (5, 2)}. (51)

From the Equations 50 and 51 we have, 4a±b =2a+b−2, i.e., 2a+2 = 0, i.e., 2a+2b = 0, which is impossible by the Lemma 4.1 but 2a+2b+2 = 0 may hold. We have this Equation when ϕ((a + b +1, 0)) = (b − a +1, 0). So 2a +2b + 2 = 0 imply a + b +1= −a − b − 1,

23 hence ϕ((−a − b − 1, 0)) = (b − a +1, 0). But from the Equations 42 and 43 we have ϕ((−a − b − 1, 0)) =(6 b − a +1, 0), which is a contradiction. Hence ϕ((3a + b, 2)) =6 (4b, 0). Case C(b)(2)(b): If ϕ((3a + b, 2)) = (3a +1, 1) then ϕ((3+ a + b, 0)) ∼ ϕ((3a + b, 2)) and ϕ((3b, 1)) = (3, 2) imply ϕ((3 + a + b, 0)) ∈{(3b + a ± 1, 0), (3+2b, 2)}. (52) From the Equations 50 and 52 we have, 3b + a ± 1=3b − a + 1, i.e., 2a =0or2a − 2 = 0, i.e., 2a − 2b = 0, which is impossible by the Lemma 4.1. Hence ϕ((3a + b, 2)) =6 (3a +1, 1). Case C(b)(2)(c): If ϕ((3a + b, 2)) = (3a − 1, 1) then ϕ((3+ a + b, 0)) ∼ ϕ((3a + b, 2)) and ϕ((3b, 1)) = (3, 2) imply ϕ((3 + a + b, 0)) ∈{(3b − a ± 1, 0), (3 − 2b, 2)}. (53) From the Equations 50 and 53 we have, 3b − a ± 1=3b − a + 1, this gives rise to two equations, out of one is impossible, namely 2 = 0. The only possibility which remains is 3b − a +1=3b − a + 1, which is an identity. So assuming this to be the case, we have ϕ((3a + b, 2)) = (3a − 1, 1) and ϕ((3 + a + b, 0)) = (3b − a +1, 0). ′ Now consider the cycle C2 : (2, 0) ∼ (3b, 1) ∼ (3a−b, 2) ∼ (3−a−b, 0) ∼ (2b−a−1, 1) ∼ ′ (2a − 1, 2) ∼ (2, 0). ϕ(C2) : (2b, 0) ∼ (3, 2) ∼ ϕ((3a − b, 2)) ∼ ϕ((3 − a − b, 0)) ∼ ϕ((2b − a − 1, 1)) ∼ (2a − b, 1) ∼ (2b, 0). ϕ((3a − b, 2)) ∼ (3, 2), ϕ((a − b, 2)) = (a +1, 1) and ϕ((2, 0)) = (2b, 0) imply ϕ((3a − b, 2)) ∈ {(3a +1, 1), (4b, 0)}. ϕ((2b − a − 1, 1)) ∼ (2a − b, 1) and ϕ((2, 0)) = (2b, 0) imply ϕ((2b − a − 1, 1)) ∈ {(2 − a ± b, 2), (2b − 2, 0)}. Now ϕ((3 − a − b, 0)) ∼ ϕ((3a − b, 2)) and ϕ((3b, 1)) = (5, 2) imply ϕ((3 − a − b, 0)) ∈{(4a ± b, 1), (5, 2), (3+2b, 2), (3b + a ± 1, 0)}. (54) Case C(b)(2)(d): If ϕ((2b−a−1, 1)) = (2b−2, 0) then ϕ((3−a−b, 0)) ∼ ϕ((2b−a−1, 1)) and ϕ((2a − 1, 2)) = (2a − b, 1) imply ϕ((3 − a − b, 0)) ∈{(2a − 3b, 1), (2 − 2a ± 1, 2)}. (55) From the Equations 54 and 55 we have, • either 2a − 3b =4a ± b, i.e., 2a +2b =0or2a +4b = 0. By Lemma 4.1 2a +2b =0 can not hold and 2a +4b = 0 can hold only if n = 9. However, direct SageMath computation for n = 9 shows that such ϕ does not exist. • or 2 − 2a ± 1 = 5, i.e., 2a + 2 = 0, i.e., 2a +2b =0or2a + 4 = 0, i.e., 4a +2b = 0. By Lemma 4.1 2a +2b = 0 can not hold and 4a +2b = 0 can hold only if n = 18. However, direct SageMath computation for n = 18 shows that such ϕ does not exist. • or 2 − 2a ± 1=3+2b, i.e., 2a +2b = 0, which is impossible by the Lemma 4.1 but 2a+2b+2 = 0 may hold. We have this Equation when ϕ((a+b+1, 0)) = (b−a+1, 0). So 2a+2b+2 = 0 imply a+b+1 = −a−b−1, hence ϕ((−a−b−1, 0)) = (b−a+1, 0). But from the Equations 42 and 43 we have ϕ((−a − b − 1, 0)) =(6 b − a +1, 0), which is a contradiction. Hence ϕ((3 − a − b, 0)) =6 (2b +3, 2).

24 Hence ϕ((2b − a − 1, 1)) =6 (2b − 2, 0). Case C(b)(2)(e): If ϕ((2b − a − 1, 1)) = (2 − a − b, 2) then ϕ((3 − a − b, 0)) ∼ ϕ((2b − a − 1, 1)) and ϕ((2a − 1, 2)) = (2a − b, 1) imply

ϕ((3 − a − b, 0)) ∈{(2a − b − 2, 1), (2b − 1 − a ± b, 0)}. (56)

From the Equations 54 and 56 we have

• either 2b − 1 − a ± b = 3b + a ± 1, i.e., 2a +2b =0or2a =0or2a + 2 = 0, i.e., 2a +2b = 0, which are impossible by the Lemma 4.1 but 2a +2b + 2 = 0 may hold. We have this Equation when ϕ((a + b +1, 0)) = (b − a +1, 0). So 2a +2b +2=0 imply a + b +1= −a − b − 1, hence ϕ((−a − b − 1, 0)) = (b − a +1, 0). But from the Equations 42 and 43 we have ϕ((−a − b − 1, 0)) =6 (b − a +1, 0), which is a contradiction. Hence ϕ((3 − a − b, 0)) =6 (3b + a +1, 0).

• or 2a − b − 2=4a ± b, i.e., 2a + 2 = 0, i.e., 2a + 2b = 0 which is impossible by the Lemma 4.1 but 2a +2b + 2 = 0 may hold. We have this Equation when ϕ((a + b +1, 0)) = (b − a +1, 0). So 2a +2b + 2 = 0 imply a + b +1= −a − b − 1, hence ϕ((−a − b − 1, 0)) = (b − a +1, 0). But from the Equations 42 and 43 we have ϕ((−a − b − 1, 0)) =6 (b − a +1, 0), which is a contradiction, so ϕ((3 − a − b, 0)) =6 (4a + b, 1).

Hence ϕ((2b − a − 1, 1)) =6 (2 − a − b, 2). Case C(b)(2)(f): If ϕ((2b − a − 1, 1)) = (2 − a + b, 2) then ϕ((3 − a − b, 0)) ∼ ϕ((2b − a − 1, 1)) and ϕ((2a − 1, 2)) = (2a − b, 1) imply

ϕ((3 − a − b, 0)) ∈{(2a − b +2, 1), (2b − 1+ a ± b, 0)}. (57)

From the Equations 54 and 57 we have,

• either 2a − b +2=4a ± b, i.e., 2a − 2 = 0, i.e., 2a − 2b =0or2a +2b − 2 = 0 which are impossible by the Lemma 4.1.

• or 2b−1+a±b =3b+a±1, this gives rise to four equations, out of three are impossible by the Lemma 4.1, namely 2 = 0, 2b = 0, 2b + 2 = 0, i.e., 2a +2b = 0. The only possibility which remains is 3b + a − 1=3b + a − 1, which is an identity. So assuming this to be the case, we have ϕ((3 − a − b, 0)) = (3b + a − 1, 0), ϕ((2b − a − 1, 1)) = (2 − a + b, 2) and ϕ((3a − b, 2)) = (3a +1, 1).

Now ϕ((4, 0)) ∼ ϕ((3b, 1)) = (3, 2), ϕ((2, 0)) = (2b, 0), ϕ((3a + b, 2)) = (3a − 1, 1) and ϕ((3a − b, 2)) = (3a +1, 1) imply ϕ((4, 0)) = (4b, 0). ϕ((2b, 0)) ∼ ϕ((1, 2)) = (b, 1), ϕ((0, 0)) = (0, 0), ϕ((a +1, 1)) = (a − b, 2) and ϕ((a − 1, 1)) = (a + b, 2) imply ϕ((2b, 0)) = (2, 0). Proceeding this way we can show that ϕ((2k, 0)) = (2kb, 0), forall k ∈ Z. So we have ϕ((2b, 0)) = (2a, 0), where k = a, which is a contradiction as we have shown that

25 ϕ((2b, 0)) = (2, 0) and 2a =6 2 by Lemma 4.1. Therefore we have ϕ((a +1, 1)) =(6 a − b, 2) and ϕ((a − 1, 1)) =(6 a + b, 2) . Case D: Let ϕ((a +1, 1)) = (a + b, 2), ϕ((a + b +1, 0)) = (a + b +1, 0), ϕ((a + b, 2)) = (a+1, 1), ϕ((a−1, 1)) = (a−b, 2), ϕ((b−a+1, 0)) = (b−a+1, 0), ϕ((a−b, 2)) = (a−1, 1). Now ϕ((a + b, 2)) = (a +1, 1), ϕ((a − b, 2)) = (a − 1, 1) and ϕ((2, 0)) ∼ ϕ((b, 1)) = (1, 2) imply ϕ((2, 0)) = (2b, 0). Consider the cycle C1 : (2, 0) ∼ (b, 1) ∼ (a + b, 2) ∼ (a + b +1, 0) ∼ (2b + a +1, 1) ∼ (2a +1, 2) ∼ (2, 0). then ϕ(C1):(2b, 0) ∼ (1, 2) ∼ (a +1, 1) ∼ (a + b +1, 0) ∼ ϕ((2b + a + 1, 1)) ∼ ϕ((2a +1, 2)) ∼ (2b, 0). ϕ((2b + a +1, 1)) ∼ (a + b +1, 0), ϕ((a +1, 1)) = (a + b, 2) and ϕ((a + b, 2)) = (a +1, 1) imply ϕ((2b + a +1, 1)) ∈{(a +1+2b, 1), (2+ b + a, 2)}. Now ϕ((2a +1, 2)) ∼ (2b, 0) and ϕ((b, 1)) = (1, 2) imply ϕ((2a +1, 2)) ∈{(2a ± b, 1), (3, 2)}. (58) Case D(a): If ϕ((2b + a +1, 1)) = (a +1+2b, 1) then ϕ((2b + a +1, 1)) ∼ ϕ((2a +1, 2)) and ϕ((a + b +1, 0)) = (a + b +1, 0) imply ϕ((2a +1, 2)) ∈{(1 + b +2a ± b, 2), (b + a +3, 0)}. (59) then From the Equation 58 and 59 we have, 1+b+2a±b = 3, i.e., 2a−2 = 0, i.e., 2a−2b =0 or 2a +2b − 2 = 0, which are impossible by the Lemma 4.1. Hence ϕ((2b + a +1, 1)) =6 (a +1+2b, 1). Case D(b): If ϕ((2b + a +1, 1)) = (2+ b + a, 2) then ϕ((2b + a +1, 1)) ∼ ϕ((2a +1, 2)) and ϕ((a + b +1, 0)) = (a + b +1, 0) imply ϕ((2a +1, 2)) ∈{(2a +1+ b ± 1, 1), (3b + a +1, 0)}. (60) From the Equation 58 and 59 we have, 2a+1+b±1=2a±b, This gives rise four equations out of which three are impossible by Lemma 4.1, namely 2 = 0, 2b = 0, 2b + 2 = 0, i.e., 2a +2b = 0. The only possiblity which remains is 2a + b = 2a + b, which is an identity. So asssuming this to be the case we have ϕ((2a +1, 2)) = (2a + b, 1), ϕ((2b + a +1, 1)) = (2 + b + a, 2). ′ Now consider the cycle C1 : (2, 0) ∼ (b, 1) ∼ (a + b, 2) ∼ (1 + a − b, 0) ∼ (2b + ′ 1 − a, 1) ∼ (2a − 1, 2) ∼ (2, 0). Then ϕ(C1) : (2b, 0) ∼ (1, 2) ∼ (a +1, 1) ∼ ϕ((1 + a − b, 0)) ∼ ϕ((2a + 1 − a, 1)) ∼ ϕ((2a − 1, 2)). ϕ((2a − 1, 2)) ∼ (2b, 0), ϕ((b, 1)) = (1, 2) and ϕ((2a +1, 2)) = (2a + b, 1) imply ϕ((2a − 1, 2)) ∈ {(2a − b, 1), (3, 2)}. Now ϕ((1 + a − b, 0)) ∼ (a +1, 1), ϕ((a + b +1, 0)) = (a + b +1, 0) and ϕ((b, 1)) = (1, 2) imply ϕ((1 + a − b, 0)) ∈ {(b + a − 1, 0), (1+2b, 2)}. ϕ((2b +1 − a, 1)) ∼ ϕ((1 + a − b, 0)) and ϕ((a + b, 2)) = (a +1, 1) imply ϕ((2b +1 − a, 1)) ∈{(a +1 − 2b, 1), (1 + b − a ± 1, 2), (a +3, 1), (b +2a ± b, 0)}. (61) Case D(b)(1): If ϕ((2a − 1, 2)) = (3, 2) then ϕ((2b +1 − a, 1)) ∼ ϕ((2a − 1, 2)) and ϕ((2, 0)) = (2b, 0) imply ϕ((2b +1 − a, 1)) ∈{(4b, 0), (3a ± 1, 1)}. (62) From the Equations 61 and 62 we have,

26 • either 4b = b +2a ± b, i.e., 2a − 2b =0or2a − 4b = 0, which are impossible by the Lemma 4.1.

• 3a ± 1= a +1 − 2b, i.e., 2a +2b =0or2a +2b − 2 = 0, which are impossible by the Lemma 4.1.

• 3a ± 1 = a + 3, i.e., 2a − 2 = 0, i.e., 2a − 2b =0or2a = 4, i.e., 4a − 2b = 0. Now 2a − 2b = 0 s impossible by the Lemma 4.1 but 4a − 2b = 0 holds only if n =7, 14. Thus ϕ((2a − 1, 2)) = (3, 2) only if n =7, 14. Case D(b)(2): If ϕ((2a − 1, 2)) = (2a − b, 1) then ϕ((2b +1 − a, 1)) ∼ ϕ((2a − 1, 2)) and ϕ((2, 0)) = (2b, 0) imply

ϕ((2b +1 − a, 1)) ∈{(2b − 2, 0), (2 − a ± b, 2)}. (63)

From the Equations 61 and 63 we have, • either 2b − 2= b +2a ± b, i.e., 2a + 2 = 0, i.e., 2a − 2b + 2 = 0, which are impossible by the Lemma 4.1.

• or 2 − a ± b =1+ b − a ± 1, This gives rise four equations out of which three are impossible by Lemma 4.1, namely 2 = 0, 2b =0,2b−2 = 0, i.e., 2a−2b = 0. The only possiblity which remains is 2 − a + b =2+ b − a, which is an identity. So asssuming this to be the case we have ϕ((2b+1−a, 0)) = (2+b−a, 0), ϕ((2a−1, 2)) = (2a−b, 1) and ϕ((1 + a − b, 0)) = (b + a − 1, 0). Now ϕ((3b, 1)) ∼ ϕ((2, 0)) = (2b, 0), ϕ((2a +1, 2)) = (2a + b, 1) and ϕ((2a − 1, 2)) = (2a − b, 1) imply ϕ((3b, 1)) = (3, 2). Consider the cycle C2 : (2, 0) ∼ (3b, 1) ∼ (3a + b, 2) ∼ (3+ a + b, 0) ∼ (2b + a +1, 1) ∼ (2a +1, 2) ∼ (2, 0). Then ϕ(C2):(2b, 0) ∼ (3, 2) ∼ ϕ((3a + b, 2) ∼ ϕ((3+ a + b, 0)) ∼ (a + b+2, 2) ∼ (2a+b, 1) ∼ (2b, 0). ϕ((3+a+b, 0)) ∼ (a+b+2, 2), ϕ((a+b+1, 0)) = (a+b+1, 0) and ϕ((2a +1, 2)) = (2a + b, 1) imply ϕ((3 + a + b, 0)) ∈{(2a + b +2, 1), (3b + a +1, 0)}. ϕ((3a + b, 2)) ∼ (3, 2) and ϕ((2, 0)) = (2b, 0) imply

ϕ((3a + b, 2)) ∈{(4b, 0), (3a ± 1, 1)}. (64)

Case D(b)(2)(a): If ϕ((3+a+b, 0)) = (2+b+2a, 1) then ϕ((3a+b, 2) ∼ ϕ((3+a+b, 0)) and ϕ((2b + a +1, 1)) = (a + b +2, 2) imply

ϕ((3a + b, 2)) ∈{(2a +1+2b ± 1, 0), (3b + a +2, 2)}. (65)

From 64 and 65 we have, 2a +1+2b ± 1=4b, i.e., 2a − 2b =0or2a − 2b + 2 = 0, which are impossible by Lemma 4.1. Hence If ϕ((3 + a + b, 0)) =(2+6 b +2a, 1). Case D(b)(2)(b): If ϕ((3+a+b, 0))= (a+1+3b, 0) then ϕ((3a+b, 2) ∼ ϕ((3+a+b, 0)) and ϕ((2b + a +1, 1)) = (a + b +2, 2) imply

ϕ((3a + b, 2)) ∈{(1 + b +3a ± b, 1), (b + a +4, 2)}. (66)

27 From 64 and 66 we have, 1+b+3a±b =3a±1, this give rise to four equations, out of three are impossible by the Lemma 4.1, namely 2 = 0, 2b =0 and 2b + 2 = 0, i.e., 2a +2b = 0. The only possibility which remains is 1 + 3a = 3a + 1, which is an identity. So assuming to be the case, we have ϕ((3a + b, 2)) = (3a +1, 1) and ϕ((3 + a + b, 0)) = (3b + a +1, 0). ′ Now consider the cycle C2 : (2, 0) ∼ (3b, 1) ∼ (3a − b, 2) ∼ (3 − a + b, 0) ∼ (2b − 1+ ′ a, 1) ∼ (2a +1, 2). Then ϕ(C2) : (2b, 0) ∼ (3, 2) ∼ ϕ((3a − b, 2)) ∼ ϕ((3 − a + b, 0)) ∼ ϕ((2b−1+a, 1)) ∼ (2a+b, 1) ∼ (2b, 0). ϕ((3a−b, 2)) ∼ (3, 2), ϕ((a−b, 2)) = (a−1, 1) and ϕ((2, 0)) = (2b, 0) imply ϕ((3a−b, 2)) ∈{(3a−1, 1), (4b, 0)}. ϕ((2b−1+a, 1)) ∼ (2a+b, 1), ϕ((2b + a + 1, 1)) = (a + b + 2, 2) and ϕ((2, 0)) = (2b, 0) imply ϕ((2b − 1 + a, 1)) ∈ {(2 + a − b, 2), (2b +2, 0)}. Now ϕ((3 − a + b, 0)) ∼ ϕ((3a − b, 2)) and ϕ((3b, 1)) = (3, 2) imply ϕ((3 − a + b, 0)) ∈{(3b − a +1, 0), (3 − 2b, 2), (5, 2), (4a ± b, 1)}. (67) Case D(b)(2)(c): If ϕ((2b−1+a, 1)) = (2b+2, 0) then ϕ((3−a+b, 0)) ∼ ϕ((2b−1+a, 1)), ϕ((3b, 1)) = (3, 2) and ϕ((2a +1, 2)) = (2a + b, 1) imply

ϕ((3 − a + b, 0)) ∈{(2a +3b, 1), (3+2a, 2)}. (68)

From the Equations 67 and 68 we have, • either 2a +3b = 4a ± b, i.e., 2a − 2b =0or2a − 4b = 0, which are impossible by lemma 4.1.

• or 3+2a =3 − 2b, i.e., 2a +2b = 0, which is impossible by Lemma 4.1.

• or 3+2a = 5, i.e., 2a − 2 = 0, i.e., 2a − 2b = 0, which is impossible by Lemma 4.1. Hence ϕ((2b − 1+ a, 1)) =6 (2b +2, 0). Case D(b)(2)(d): If ϕ((2b − 1+ a, 1)) = (2+ a − b, 2) then ϕ((3 − a + b, 0)) ∼ ϕ((2b − 1+ a, 1)), ϕ((b − a +1, 0)) = (b − a +1, 0) and ϕ((2a +1, 2)) = (2a + b, 1) imply

ϕ((3 − a + b, 0)) ∈{(3b +1 − a, 0), (2a + b − 2, 1)}. (69)

From the Equations 67 and 69 we have, • either 2a + b − 2=4a ± b, i.e., 2a + 2 = 0, i.e., 2a +2b =0or2a − 2b + 2 = 0, which are impossible by lemma 4.1.

• 3b +1 − a = 3b − a ± 1, this give rise to two equations, out of one is impossible, namely 2 = 0. The only possibility which remains is 3b +1 − a =3b − a + 1, which is an identity. So assuming to be the case, we have, ϕ((3 − a + b, 0)) = (3b − a +1, 0), ϕ((3a − b, 2)) = (3a − 1, 1) and ϕ((2b + a − 1, 1)) = (2+ a − b, 2). Now ϕ((4, 0)) ∼ ϕ((3b, 1)) = (3, 2), ϕ((2, 0)) = (2b, 0), ϕ((3a + b, 2)) = (3a +1, 1) and ϕ((3a − b, 2)) = (3a − 1, 1) imply ϕ((4, 0)) = (4b, 0). ϕ((2b, 0)) ∼ ϕ((1, 2)) = (b, 1), ϕ((0, 0)) = (0, 0), ϕ((a +1, 1)) = (a + b, 2) and ϕ((a − 1, 1)) = (a − b, 2) imply ϕ((2b, 0)) = (2, 0).

28 Proceeding this way we can show that ϕ((2k, 0)) = (2kb, 0), forall k ∈ Z. So we have ϕ((2b, 0)) = (2a, 0), where k = a, which is a contradiction as we have shown that ϕ((2b, 0)) = (2, 0) and 2a =6 2 by Lemma 4.1. Therefore from the case D(b)(1) we have ϕ((a +1, 1)) = (a + b, 2) and ϕ((a − 1, 1)) = (a − b, 2) only if n = 7, 14. This completes the proof. Now, the proof of Theorem 3.4 follows from Theorem 4.1, Theorem 4.2 and Theorem 4.3.

5. Open Issues In this paper, we introduced an inﬁnite family of half-transitive Cayley graphs. How- ever, a few issues are still pending and can be topics for further research. 1. Full Automorphism Group: It was shown that hα,β,γi is a subgroup of the full automorphism group. It remains to be shown (as observed in SageMath) that G = hα,β,γi for n =76 , 14.

2. Structural Properties of Γ(n, a): We have computed the girth for some special values of n and shown that Γ(n, a) is Hamiltonian if n is odd. However, the girth and Hamiltonicity for general values of n are still unanswered. Similarly, other structural properties like diameter, domination number are few open issues.

Acknowledgement The second author acknowledge the funding of DST-SERB-SRG Sanction no. SRG/2019/ 000475, Govt. of India.

References [1] B. Alspach, D. Marusic and L. Nowitz: Constructing graphs which are 1/2-transitive, Journal of the Australian Mathematical Society, 56(3), pp. 391-402, 1994.

[2] I. Z. Bouwer: Vertex and edge-transitive but not 1-transitive graphs, Canad. Math. Bull. 13, (1970), 231-237.

[3] J.Chen, C.H. Li and Akos´ Seress: A family of half-transitive graphs, Electronic Journal of Combinatorics, 20(1), P56, 2013.

[4] H. Cheng and L. Cui: Tetravalent half-arc-transitive graphs of order p5, Applied Math- ematics and Computation, 332: 506-518, 2018.

[5] Y.Q. Feng, J.H. Kwak, X. Wang and J.X. Zhou: Tetravalent half-arc-transitive graphs of order 2pq, Journal of Algebraic Combinatorics, 33: 543-553, 2011.

[6] Y.Q. Feng, J.H. Kwak, M.Y. Xu and J.X. Zhou: Tetravalent half-arc-transitive graphs of order p4, European Journal of Combinatorics 29(3): 555-567, 2008.

29 [7] Y.Q. Feng, K. Wang and C. Zhou: Tetravalent half-arc-transitive graphs of order 4p, European Journal of Combinatorics, 28: 726-733, 2007. [8] C. Godsil and G.F. Royle: Algebraic Graph Theory, Graduate Texts in Mathematics, 207, Springer-Verlag, 2001. [9] D.F. Holt: A Graph Which Is Edge Transitive But Not Arc Transitive, Journal of Graph Thery, 5, 201-204, 1981. [10] D. Marusic: Hamiltonian Circuits in Cayley Graphs, Discrete Mathematics, 46: 49-54, 1983. [11] W. Stein and others: Sage Mathematics Software (Version 7.3), Release Date: 04.08.2016, http://www.sagemath.org. [12] W.T. Tutte: Connectivity in Graphs, Univ. of Toronto Press, Toronto, 1966. [13] C. Zhou and Y.Q. Feng: An inﬁnite family of tetravalent half-arc-transitive graphs, Discrete Mathematics, 306: 2205-2211, 2006.

Appendix: Sage Code for Γ(n,a) for n = 7,a = 2 n=7 a=2 b=mod(a^2,n) A=list(var(’A_%d’ % i) for i in range(n)) B=list(var(’B_%d’ % i) for i in range(3)) C = cartesian_product([A, B]) V=C.list() E=[] Gamma=Graph() Gamma.add_vertices(V) for i in range(n): for j in range(3): E.append(((A[i],B[j]),(A[mod(a*i+1,n)],B[mod(j-1,3)]))) E.append(((A[i],B[j]),(A[mod(a*i-1,n)],B[mod(j-1,3)]))) E.append(((A[i],B[j]),(A[mod(b*i+b,n)],B[mod(j+1,3)]))) E.append(((A[i],B[j]),(A[mod(b*i-b,n)],B[mod(j+1,3)]))) Gamma.add_edges(E) G=Gamma.automorphism_group() for f in G: if f((A[0],B[0]))==(A[0],B[0]) and f((A[b],B[1]))==(A[1],B[2]) and f((A[1],B[2]))==(A[n-1],B[2]): print "sucess"